IN  MEMORIAM 
FLOR1AN  CAJOR1 


SECOND  COURSE   IN   ALGEBRA 


A  SERIES  OF  MATHEMATICAL  TEXTS 

EDITED   BY 

EARLE  RAYMOND  HEDRICK 


THE  CALCULUS 

By    ELLERT    WILLIAMS    DAVIS    and    WILLIAM    CHARLES 
BRENKE. 

ANALYTIC   GEOMETRY  AND   ALGEBRA 

By  ALEXANDER  ZIWET  and  Louis  ALLEN  HOPKINS. 
ELEMENTS   OF  ANALYTIC   GEOMETRY 

By  ALEXANDER  ZIWET  and  Louis  ALLEN  HOPKINS. 
PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
COMPLETE   TABLES 

By  ALFRED  MONROE  KENYON  and  Louis  INGOLD. 
PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
BRIEF  TABLES 

By  ALFRED  MONROE  KENYON  and  Louis  INGOLD. 
ELEMENTARY   MATHEMATICAL   ANALYSIS 

By  JOHN  WESLEY  YOUNG  and  FRANK  MILLETT  MORGAN. 
PLANE   TRIGONOMETRY 

By  JOHN  WESLEY  YOUNG  and  FRANK  MILLETT  MORGAN. 
COLLEGE  ALGEBRA 

By  ERNEST  BROWN  SKINNER. 

ELEMENTS  OF  PLANE   TRIGONOMETRY  WITH  COM- 
PLETE TABLES 

By  ALFRED  MONROE  KENYON  and  Louis  INGOLD. 
ELEMENTS  OF  PLANE  TRIGONOMETRY  WITH  BRIEF 
TABLES 

By  ALFRED  MONROE  KENYON  and  Louis  INGOLD. 

THE   MACMILLAN  TABLES 

Prepared  under  the  direction  of  EARLE  RAYMOND  HEDRICK. 
PLANE  GEOMETRY 

By  WALTER  BURTON  FORD  and  CHARLES  AMMERMAN. 
PLANE   AND   SOLID   GEOMETRY 

By  WALTER  BURTON  FORD  and  CHARLES  AMMERMAN. 
SOLID  GEOMETRY 

By  WALTER  BURTON  FORD  and  CHARLES  AMMERMAN. 
CONSTRUCTIVE   GEOMETRY 

Prepared  under  the  direction  of  EARLE  RAYMOND  HEDRICK. 
JUNIOR   HIGH   SCHOOL   MATHEMATICS 

By  W.  L.  VOSBURGH  and  F.  W.  GENTLEMAN. 


This  book  is  issued  in  a  form  identical  with  that  of  the  books  announced  above, 
as  is  the  First  Course  in  Algebra  by  the  same  authors. 


SECOND  COURSE  IN  ALGEBRA 


BY 

WALTER  BURTON  FORD 

n 

PROFESSOR   OF    MATHEMATICS,    THE    UNIVERSITY    OF    MICHIGAN 
AND 

CHARLES  AMMERMAN 

THE    WILLIAM    McKINLEY   HIGH    SCHOOL,    ST.    LOUIS 


gorfe 

THE  MACMILLAN   COMPANY 

1920 

All  rights  reserved 


COPYRIGHT,  1920, 
BY  THE  MACMILLAN  COMPANY. 

Set  up  and  electrotyped.     Published  January,  1920 


Nortoooti 

J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE 

IN  the  present  volume,  which  is  intended  as  a  sequel  to 
the  author's  First  Course  in  Algebra,  the  following  features 
may  be  noted : 

(a)  The  first  seven  chapters  furnish  a  systematic  review 
of  the  ordinary  first  course  up  to  and  including  the  subject 
of  simultaneous  linear  equations.  In  these  opening  chapters 
the  aim  is  to  state  briefly  and  concisely  those  fundamental 
principles  which  are  basal  in  all  algebraic  work  and  with 
which  the  pupil  already  has  some  acquaintance.  The 
method  of  treatment  is  largely  inductive,  being  based  upon 
solved  problems  and  other  illustrative  material  rather  than 
upon  any  attempt  at  proofs  or  formal  demonstrations.  The 
various  principles  are  explicitly  stated,  however,  at  all  points 
in  the  form  of  rules,  which  are  clearly  set  off  in  italics.  Upon 
this  plan  the  pupil  is  rapidly  and  effectively  prepared  for  un- 
dertaking the  newer  and  more  advanced  topics  which  follow. 

(6)  Chapters  VIII-IX  (Square  Root  and  Radicals)  are 
essentially  a  reproduction  of  the  corresponding  chapters  in 
the  First  Course,  but  all  of  the  problems  are  new.  This  is 
true  also  of  the  early  parts  of  Chapter  X  (Quadratic  Equa- 
tions). These  are  topics  which  usually  present  more  than 
average  difficulty ;  hence,  even  for  pupils  who  have  studied 
them  carefully  before,  a  complete  treatment  of  them  is  de- 
sirable in  the  second  course. 

The  introduction  and  use  of  tables  of  square  and  cube 
roots  at  this  point  (§  43)  is  to  be  especially  noted.  It  seems 


vi  PREFACE 

clear  that  pupils  should  be  made  familiar  with  such  tables 
at  an  earlier  date  than  formerly.  One  strong  reason  for  this 
is  that  a  constantly  increasing  number  of  students  pass 
directly  from  the  high  schools  into  technical  pursuits  where 
facility  in  the  manipulation  of  tables  of  all  kinds  is  especially 
desirable. 

(c)  Part  II,  comprising  Chapters  XI-XX,  presents  the  usual 
topics  of  the  advanced  course.     The  order  of  arrangement 
follows,  so  far  as  possible,  that  of  the  difficulty  of  the  various 
subjects,  and  the  whole  has  been  prepared  with  a  view  of 
introducing  a  relatively  large  number  of  simple  illustrative 
examples  drawn  from  nature   and   the  arts.     Throughout 
the  development,  however,  due  emphasis  has  been  given 
to  those  fundamental  disciplinary  values  which  should  be 
preserved  in  any  course  in  mathematics. 

Among  the  unusual  features,  it  may  be  observed  that  the 
detailed  consideration  of  exponents  and  radicals  has  been 
delayed  until  logarithms  are  about  to  be  taken  up.  These 
topics  in  their  extended  sense  have,  in  fact,  but  little  to  do 
with  algebra  until  that  time. 

Again,  the  chapter  on  logarithms  is  unusually  full  and  com- 
plete. All  the  essential  features  of  this  relatively  difficult 
but  increasingly  important  subject  are  presented  in  detail. 
In  the  past,  much  has  ordinarily  been  left  for  the  teacher  to 
explain. 

(d)  Functions,    Mathematical    Induction    (including  the 
proof  of  the  Binomial  Theorem),  and  Determinants  have 
been  grouped  together  under  the  title  Supplementary  Topics. 
In  fact,  these  subjects  lie  on  the  border  line  between  the 
second  course  and  the  college  course.     Only  the  elements  of 
each  are  taken  up,  but  there  is  enough  to  show  its  im- 
portant bearing  in   algebra  and  to  pave  the  way  for  its 
further  development  in  the  college  course.     For  example, 


PREFACE  vii 

the  study  of  functions  is  so  presented  that  it  at  once  amplifies 
material  to  be  found  in  the  earlier  chapters  of  the  book  and 
brings  in  new  material  which  is  connected  with  the  graphical 
study  of  the  theory  of  equations.  Thus  it  serves  as  an  in- 
troduction to  the  latter  subject  as  presented  in  the  usual 
college  texts. 

As  in  the  authors'  other  texts,  a  star  (*)  has  been  placed 
against  certain  sections  that  may  be  omitted  if  desired  with- 
out destroying  the  continuity  of  the  whole. 

WALTER  B.  FORD. 
CHARLES  AMMERMAN. 


TABLE   OF  CONTENTS 

PART   I.     REVIEW   TOPICS 

CHAPTER  PAGE 

I.     FUNDAMENTAL  NOTIONS 1 

II.     SPECIAL  PRODUCTS  AND  FACTORING        ...  17 

III.  HIGHEST  COMMON  FACTOR  AND  LOWEST  COMMON 

MULTIPLE 26 

IV.  FRACTIONS 29 

V.     SIMPLE  EQUATIONS           .  • 36 

VI.     GRAPHICAL  STUDY  OP  EQUATIONS  ....  41 

VII.     SIMULTANEOUS  EQUATIONS  SOLVED  BY  ELIMINATION  50 

VIII.     SQUARE  ROOT 59 

IX.     RADICALS 67 

X.     QUADRATIC  EQUATIONS 78 

PART   II.    ADVANCED   TOPICS 

XI.     LITERAL  EQUATIONS  AND  FORMULAS       ...  95 

XII.     GENERAL  PROPERTIES  OF  QUADRATIC  EQUATIONS  .  108 

XIII.  IMAGINARY  NUMBERS 116 

XIV.  SIMULTANEOUS  QUADRATIC  EQUATIONS  .         .         .  122 
XV.     PROGRESSION 141 

XVI.     RATIO  AND  PROPORTION 166 

XVII.     VARIATION        .         .         .         .        .        .        .         .178 

XVIII.     EXPONENTS 193 

XIX.     RADICALS 205 

XX.     LOGARITHMS 216 

viii 


TABLE   OF  CONTENTS  ix 


PART   III.     SUPPLEMENTARY   TOPICS 

CHAPTER  PAGE 

XXI.     FUNCTIONS       .                 .                 ....  245 

XXII.     MATHEMATICAL  INDUCTION  —  BINOMIAL  THEOREM  255 

XXIII.     THE  SOLUTION  OF  EQUATIONS  BY  DETERMINANTS  265 


PART   IV.     TABLES 

TABLE  OF  POWERS  AND  ROOTS 275 

TABLE  OF  IMPORTANT  NUMBERS 296 

INDEX  .        .        .297 


LAPLACE 

(Pierre  Simon  Laplace,  1749-1827) 

Famous  in  mathematics  for  his  researches,  which  were  of  a  most  advanced 
kind,  and  especially  famous  in  astronomy  for  his  enunciation  of  the  Nebular 
Hypothesis.  Interested  also  in  physics  and  at  various  times  held  high  polit- 
ical offices  under  Napoleon. 


SECOND  COURSE  IN  ALGEBRA 

PART  I.    REVIEW  TOPICS  f 

CHAPTER  I 
FUNDAMENTAL   NOTIONS 

1.  Negative  Numbers.  In  the  First  Course  in  Algebra 
it  was  shown  how  negative  as  well  as  positive  numbers  may 
be  used,  the  one  being  quite  as  common  as  the  other  in  every- 
day life. 

Thus  +15°  (or  simply  15°)  means  15°  above  0°,  while  -15° 
means  15°  below  0.  Similarly,  +$25  (or  simply  $25)  means  a  gain 
or  asset  of  $25,  while  —$25  means  a  loss  or  debt  of  the  same  amount. 

Negative  numbers  are  compared  with  each  other  in  much 
the  same  way  as  positive  numbers.  Thus,  just  as  in  arith- 
metic 4  is  less  than  5,  3  is  less  than  4,  etc.,  until  finally  we 
say  that  0  is  less  than  1,  so  we  continue  this  idea  in  algebra 
by  saying  that  —  1  is  less  than  0,  —2  is  less  than  —  1,  etc. 

The  whole  situation  regarding  the  size  of  numbers  is 
vividly  brought  out  to  the  eye  in  the  figure  below : 

I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    i    I    II    I    |    I    I    I 

-12-11-10-9-8—7-6-5-4-3-2-1     0      1     2     3     4~  ~S     €    1     8     9    10  11    12 

FIG.  1. 

Here  the  positive  (+)  numbers  are  placed  in  their  order  to 
the  right  of  the  point  marked  0,  while  the  negative  (  — ) 
numbers  are  placed  in  their  order  to  the  left  of  that  same 
point.  This  figure  shows  all  numbers  (positive  and  negative) 
arranged  in  their  increasing  order  from  left  to  right. 

t  Chapters  I-X  (pp.  1-94)  furnish  a  review  of  the  First  Course, 
The  remaining  chapters  deal  with  more  advanced  topics. 
B  1 


2  SECOND  COURSE  IN  ALGEBRA  [I,  §1 

In  Fig.  1,  only  the  positive  and  negative  integers  and  zero 
are  actually  marked.  A  complete  figure  would  show  also 
the  positions  of  the  fractions.  Thus  £  is  located  at  the  point 
halfway  between  the  0  and  the  1 ;  2^  is  located  at  the  point 
one  third  the  way  from  2  to  3 ;  —  f  is  at  the  point  f  the  dis- 
tance from  0  to  —  1 ;  —  5£  is  at  the  point  f  the  distance  from 
—  5  to  —  6 ;  and  so  on  for  all  fractions. 

By  the  numerical  value  (or  absolute  value)  of  a  negative 
number  is  meant  its  corresponding  positive  value. 

Thus  the  numerical,  or  absolute,  value  of  —3  is  +3,  or  simply  3. 

NOTE.  The  numerical  value  of  any  positive  number  is  the 
number  itself. 

EXERCISES 

In  each  of  the  following  exercises,  state  which  of  the  two 
numbers  is  the  larger.  First  locate  the  number  at  its  proper 
place  on  the  line  shown  in  §  1. 

1.  7,  10.  6.  |,  f.  9.  3i,  If. 

2.  7,  -10.  6.  f,-f.  10.  -3*,  -If. 

3.  -7,  10.  7.  -f,  f.  11.  -i  0. 

4.  -7,  -10.  8.  -f,  -f.  12.  -.3,  -.05 

2.  Operations  with  Numbers.  The  following  facts  will 
be  recalled  from  the  First  Course  in  Algebra^ 

(a)  To  add  two  numbers  having  like  signs,  add  their  absolute 
values  (§1)  and  prefix  the  common  sign. 

Thus  (-5)+(-6)  =  -ll. 

(b)  To  add  two  numbers  having  unlike  signs,  find  the  dif- 
ference between  their  absolute  values  and  prefix  the  sign  of  the 
one  whose  absolute  value  is  the  greater. 

Thus  (+3)+(-5)  =  -2. 

t  References  to  the  authors'  First  Course  in  Algebra  are  given  in 
this  book  by  page  number. 


I,  §2]  FUNDAMENTAL  NOTIONS  3 

NOTE.  If  we  have  more  than  two  numbers  to  add  together,  as 
for  example  (+4)+(-7)+(+5)+(-6)+(-l),  the  customary 
way  is  to  add  all  the  positive  parts  together,  then  all  the  negative 
parts,  and  finally  to  add  the  two  results  thus  obtained.  Thus,  in 
the  example  just  mentioned,  the  sum  of  the  positive  parts  is  4  +5  =  9, 
while  the  sum  of  the  negative  parts  is  (—  7)  +  (  —  6)  +(  —  1)  =  — 14. 
The  final  result  sought  is,  therefore,  (+9)  +( -14)  =  -5. 

(c)  To  subtract  one  number  from  another,  change  the  sign 
of  the  subtrahend  and  add  the  result  to  the  minuend. 

Thus  (-7)  -  (-5)  =  (-7)+(+5)  =  -2. 

(d)  To  multiply  one  number  by  another,  find  the  product 
of  their  absolute  values,  and  take  it  positive  if  the  two  numbers 
have  the  same  sign,  but  negative  if  they  have  unlike  signs. 

Thus  (+3)  •  (+2)  =  +6,  and  (-3)  •  (-2)  =  +6;  but  (-3)  •  (+2)  = 
-6  and  (+3)  •  (-2)  =  -6. 

(e)  To  divide  one  number  by  another,  find  the  quotient  of 
their  absolute  values,  and  take  it  positive  if  the  numbers  have 
the  same  sign,  but  negative  if  they  have  unlike  signs. 

Thus  (+8H(+2)  =  +4,  and  (-8)-K-2)  =  +4;  but  (-8)^(+2)  = 
-4,  and  (+8)- ( -2)  =  -4. 

EXERCISES 

Determine  the  value  of  each  of  the  following  indicated  ex- 
pressions. 

1.  (+4)  +  (+7).  8.  (+25)  +  (-32)  +  (- 

2.  (-4)  +  (+7).  9.  (+7) -(+4). 

3.  (+4)  +  (-7).  10.  (+7)-(-4). 

4.  (-4)  +  (-7).  11.  (-7)-(+4). 

5.  (+9)  +  (-27).  12.  (-7)-(-4). 

6.  (-32)  +  (+16).  13.  (+34) -(-63). 

7.  (-3)  +  +2)  +  (-l).  14.  -54--32. 


SECOND  COURSE   IN  ALGEBRA  [I,  §2 


16. 

[HINT.     By  (c)    of  §  2,  this  may  first  be  changed  into  the  form 

(+6)  +(+4)  +(-2)  +(+5).] 

16. 

17.    (-23)  +  (+32)-(- 

18.  (+3)  -(-4).  22.  (+24)  -5-  (-6). 

19.  (-4)  -(+3).  23.   (-36)  -5-  (+6). 

20.  (+5)  •  (+4).  24.  (-55)  -(-11). 

21.  (-5)  -(-4).  25.   (f)-(-i). 

3.  Use  of  Letters  in  Algebra.  Algebra  is  distinguished 
from  arithmetic  not  only  because  of  its  use  of  negative  num- 
bers, but  also  because  of  its  general  use  of  letters  to  represent 
numbers.  This  is  useful  in  many  ways.  In  particular,  it 
enables  us  to  solve  problems  in  arithmetic  which  would  other- 
wise be  very  difficult.  The  following  facts  and  definitions 
will  be  recalled  from  the  First  Course  in  this  connection. 

The  sum  of  any  two  numbers,  as  x  and  y,  is  represented  by 
x+y. 

The  difference  between  any  two  numbers,  as  x  and  y 
(meaning  the  number  which  added  to  y  gives  x),  is  represented 
bys-y. 

The  product  of  any  two  numbers,  such  as  x  and  y,  is  written 
in  the  form  xy.  It  has  the  same  meaning  as  xXy,  or  x  -  y. 
Either  of  the  numbers  thus  multiplied  together  is  called  a 
factor  of  the  product. 

The  quotient  of  x  divided  by  y  is  expressed  either  by  x  -5-  y, 

or  by  -,  or  by  x/y. 

The  product  x  •  x  is  represented  by  x2  and  is  read  x  square; 
similarly,  x  •  x  •  x  is  represented  by  x3  and  is  read  x  cube. 
More  generally,  x  •  x  •  x  —  to  n  factors  is  represented  by 
xn  and  is  read  x  to  the  nth  power.  The  letter  n  as  thus 
used  in  xn  is  called  the  exponent  of  x. 


I,  §  3]  FUNDAMENTAL  NOTIONS  5 

The  symbol  Vx  denotes  that  number  which  when  squared 
gives  x.  It  is  called  the  square  root  of  x.  Similarly,  \/x  is 
ailed  the  cube  root  of  x  and  denotes  that  number  which 
when  cubed  gives  x.  In  general,  ~\/x  is  called  the  nth  root  of 
x,  and  denotes  that  number  which  when  raised  to  the  nth 
power  gives  x.  The  letter  n  as  thus  used  in  Vx  is  called  the 
index  of  the  root. 

Whenever  one  or  more  letters  are  combined  in  such  a  way 
as  to  require  any  of  the  processes  just  described,  the  result 
is  called  an  expression. 

Thus  2  x  +3  y,  ax-bxy,  6  ran  -3V^+2Vn,  and  2  x+yz-x*  + 
xyz  are  expressions. 

An  expression  is  read  from  left  to  right  in  the  order  in 
which  the  indicated  processes  occur.  Indicated  multiplica- 
tions and  divisions  are  to  be  carried  out,  in  general,  before 
the  indicated  additions  and  subtractions. 

Thus  2  x+y2—  x3+xyz  is  read  "Two  x  plus  y  square  minus  x 
cube  plus  xyz'1 

EXERCISES 

Read  each  of  the  following  expressions : 
1.   2x2.  6. 


6.   ±_|_± ±£ 

y    w     2 

3.  a»-2o&+6>.      7.    VS+^.  n.   x3/£±2. 

s-y 

4.  ms-n3.  8.    v^+v^.         12.    •\/aw+&r. 
13.   Express  each  of  the  following  ideas  in  letters. 

(a)  The  sum  of  the  squares  of  x  and  y. 

(b)  The  difference  between  m  cube  and  n  cube. 

(c)  Three  times  the  product  of  mn  diminished  by  twice 
the  quotient  of  x  divided  by  the  square  root  of  y. 


6  SECOND  COURSE   IN  ALGEBRA  [I,  §  3 

14.  The  fact  that  the  area  of  any  rectangle  is  equal  to  the 
product  of  its  two  dimensions  (length  and  breadth)  is  ex- 
pressed by  the  formula  A=ab.  Express  similarly  in  words 
the  meaning  of  each  of  the  following  familiar  formulas  : 

(a)  A  =|  bh.        (Formula  for  the  area  of  a  triangle.) 

(b)  A  =irr2.          (Formula  for  the  area  of  a  circle.) 

(c)  C  =  2irr.        (Formula    for    the    circumference    of    a 

circle.) 

(d)  h2  =  a?+b2.     (Theorem  of  Pythagoras  concerning  any 

right  triangle.) 

(e)  V  =  %  Tir3.       (Formula  for  the  volume  of  a  sphere.) 
(/)   A  =  4  TIT*.       (Formula  for  the  area  of  a  sphere.) 

4.  Evaluation  of  Expressions.  Whenever  the  values  of 
the  letters  in  an  expression  are  given,  the  expression  itself 
takes  on  a  definite  value.  To  obtain  this  value,  we  must 
work  out  the  values  of  the  separate  parts  of  the  expression 
and  then  combine  them  as  indicated. 

EXAMPLE.     Find  the  value  of  the  expression 
o«+2-bc-c» 

a+b+c 
when  a  =  1,  b  =  —2,  and  c  =  3. 

SOLUTION.  Giving  a,  6,  and  c  their  assigned  values,  the  expres- 
sion becomes 

l2+2-  (-2).  3-33_l+(-12)+(-27)_-38_ 

~~  :"" 


In  evaluating  expressions,  it  is  useful  to  remember  the 
following  general  facts,  which  result  from  (d)  of  §  2  : 

(a)  The  sign  of  the  product  of  an  even  number  of  negative 
factors  is  positive. 

Ex.     (-2).  (-3)-  (-1).  (-4)  =  +24. 


I,  §  4]  FUNDAMENTAL  NOTIONS  -f- 

(b)  The  sign  of  the  product  of  an  odd  number  of  negative 
factors  is  negative. 

Ex.     (-2)-  (-3)-  (-l)  =  -6. 

(c)  A  negative  number  raised  to  an  even  power  gives  a 
positive  result,  but  if  raised  to  an  odd  power  gives  a  negative 
result. 

Ex.     (  -2)4  =  +16,  but  (  -2)3  =  -8. 

(d)  An  odd  root  of  a  negative  number  is  negative. 
Ex.     ^27  =-3;    v^32=-2';   ^J=—  1. 

EXERCISES 

Evaluate  each  of  the  following  expressions  for  the  indi- 
cated values  of  the  letters. 

1.  a2+2a&+62,  when  a=l,  b=-l. 

2.  4xzy+4:xy*+xyz,  when  x=—  1,  y  =  2,  z=—  3 

3.  ?™!±Z^whenm  =  2,n  =  6. 

2  m—n 

4.  2v/2x-3v/9Y,  when  a;  =18,  2/=  -3. 
5. 


,  whenz=-l,  2/=-3,  and  a  =  l. 
,  when  m  =  2,  n=  -2,  x  =  5,  y  =  6. 

8.   By  means  of  the  formulas  in  Ex.  14,  p.  6,  find 

(a)  The  area  of  the  circle  whose  radius  is  3  feet. 

[HINT.     Take7r=3i.] 

(6)  The  circumference  of  the  circle  whose  radius  is  1^  feet. 

(c)  The  volume  of  the  sphere  whose  radius  is  5  inches. 

(d)  The  area  of  the  sphere  whose  diameter  is  1  yard. 


8  SECOND  COURSE  IN  ALGEBRA  [I,  §  5 

5.   Definitions.    A  monomial  (or  term)  is  an  expression 
not  separated  into  parts  by  the  signs  +  or  —  ,  as  5  xzy. 
A  binomial  is  an  expression  having  two  terms,  as  3  x2  —  4  yx. 
A   trinomial   is   an    expression    having   three   terms,    as 


Any  expression  containing  only  powers  of  one  or  more 
letters  is  called  a  polynomial. 

A  polynomial  is  said  to  be  arranged  in  descending  powers 
of  one  of  its  letters  if  the  term  containing  the  highest  power 
of  that  letter  is  placed  first,  the  term  containing  the  next 
lower  power  is  placed  second,  and  so  on. 

Thus,  if  we  arrange  \  x3  +-J-  x5  —  1  -\-x  —  3  x4  in  descending  powers 
of  x,  it  becomes  -J  x6  —3  x4  -f  -J-  x3  -fa;  —  1. 

Similarly,  a  polynomial  is  said  to  be  arranged  in  ascending 
powers  of  one  of  its  letters  if  the  term  containing  the  lowest 
power  of  that  letter  is  placed  first,  the  term  containing  the 
next  higher  power  is  placed  second,  and  so  on. 

Thus,  if  we  arrange  \  x3  -\-\  z6  —  1  -fa?  —  3  x4  in  ascending  powers  of 
z,  it  becomes  -l+x+%x3-3  z4+|  x6. 

Whenever  a  term  is  broken  up  into  two  factors,  either 
factor  is  known  as  the  coefficient  of  the  other  one.  Usually 
the  word  is  used  to  designate  the  factor  written  first. 

Thus,  in  4  xy,  4  is  the  coefficient  of  xy  ;  in  ax,  a  is  the  coefficient 
of  x,  etc. 

A  common  factor  of  two  or  more  terms  is  a  factor  that  oc- 
curs in  each  of  them. 

Thus  5  x,  ax,  x2,  and  x3  have  the  common  factor  x. 

Whenever  two  or  more  terms  have  a  common  literal  factor, 
they  are  said  to  be  like  terms  with  respect  to  that  factor. 

Thus  5  x,  ax,  x,  and  —  2  x  are  like  terms  with  respect  to  x  ; 
and  2a(x—y)  and  3  b(x—  y)  are  like  terms  with  respect  to  x—  y. 


I,  §  7]  FUNDAMENTAL  NOTIONS  9 

6.  Addition  and  Subtraction  of  Expressions.    The  follow- 
ing rules  will  be  recalled  from  the  First  Course,  pp.  49-57. 

(a)  To  add  like  terms,  add  the  coefficients  for  a  new  coeffi- 
cient and  multiply  the  result  by  the  common  factor. 

Thus  3  x  +5  x  -4  x  =  (3  +5  -4)z  =4x. 
Similarly,  m(x—y)+n(x—y)  =  (m+ri)(x—y}. 

(b)  To  add  polynomials,  write  like  terms  in  the  same  column, 
find  the  sum  of  the  terms  in  each  column,  and  connect  the  results 
with  the  proper  signs. 

Thus,  in  adding  3a+4b+2c,  5  a  +3  b -2  c,  and  7  a  -9  6-5  c, 
the  work  appears  as  follows : 

3a+46+2c 
5  a +3  6-2 c 
7  a-9  b-5c 
15  a—  26-5c.    Ans. 

(c)  To  subtract  a  term  from  another  like  term,  change  the 
sign  of  the  subtrahend  and  add  the  result  to  the  minuend. 

Thus  8  x*y  -  ( -3  x*y)  =8  x*y  +3  x*y  =  11  x*y. 

(d)  To  subtract  one  polynomial  from  another,  change  all 
signs  in  the  subtrahend  and  add  the  result  to  the  minuend. 

Thus,  in  subtracting  4  mn  —2  nr  +3  p  from  5  ran  —  4  nr  —  4  p, 
what  we  have  to  do  is  to  add  —4  ran  +2  nr  —  3  p  to  5  ran  —4  nr  —  4  p. 
Adding  these  (see  the  preceding  rule  for  addition  of  polynomials) 
gives  ran —2  nr  — 7  p.  Ans. 

NOTE.  If  two  or  more  expressions  can  be  arranged  according  to 
the  descending  powers  of  some  letter  (§  5),  it  is  usually  best  to  do 
so  before  attempting  to  add,  subtract,  or  perform  other  operations 
upon  them. 

7.  Parenthesis  (  ),  Bracket  [  ],  Brace  J   j,  and  Vincu- 
lum      .     These  are  symbols  for  grouping  terms  that  are  to 
be  taken  as  one  single  number  or  expression. 

Thus  4  x  —  (x  +3  y  —  z)  means  that  x  +3  y  —  z  as  a  whole  is  to  be 
subtracted  from  4  x. 


10  SECOND  COURSE   IN  ALGEBRA  [I,  §  7 

The  following  rules  will  be  recalled  : 

(a)  A  parenthesis  preceded  by  the  sign  +  (either  expressed 
or  understood]  may  always  be  removed  without  any  other  change. 

(b)  A  parenthesis  preceded  by  the  sign  --  may  be  removed 
provided  the  sign  of  each  term  in  the  parenthesis  be  first  changed. 

Thus          2  a+3  6  +  (z-3  y+z)  =  2  a+3  b+x-3  y+z 
but  2  a+3  b-(x-3y+z)=2a+3b-x+3y-z. 

EXERCISES 

1.  State  the  common  factors  in  each  of  the  following 
expressions. 

(a)  -4  x,  -5  x,  6  x.        (c)  a(z+l)2,  6(z+l),  c(;r+l)3. 

(b)  rs,  3  rs,  -  10  r2s.         (d)  2  mn(a+b),  4  m2n(a-b),  8  ran2. 

2.  Add  7a+66-3cand4a-76+4c. 

3.  Add2x+3y—  2xy,  7  xy—  4x  —  9?/,  andTx  —  5xy  —  4y. 

4.  Add     z-S-T^+lSz3,      4+14  z3-!!  x-x\      and 


[HINT.     See  Note,  §  6.] 

5.  Add  4(m+n)-3(g-r)  and  4(m+n)+6(5-r). 

6.  Add      10(a+6)-ll(6+c),      3(a+6)-5(c+d),      and 


7.  Add2mx+3nx  —  £qx,  nx+lqx—ry,  and  py  —  qz-\-3w. 

8.  Subtract  3  x  —  2  i/-fz  from  5  x  —  y-\-  z. 

9.  Subtract    4  x8  -  8  -  13  x2+  15  a;    from    6  x2+  19  x3  -  4 
+12*. 

[HINT.     See  Note,  §  6.] 

10.  From  13  a+5  6-4  c  subtract  8  a+9  6+10  c. 

11.  From  2  a+3  c+d  subtract  a  —  6+c. 

12.'  From  3  x2+7  x+  10  subtract  -x2-x-6. 

13.  Subtract  l-a+a2-a3  from  1-a3. 

14.  From  the  sum  of  x2  —  4  xy+y*  and  6  x2  —  2  xy+3  y2  sub- 
tract 3  a;2  —  5xy+7  y2.     Do  it  all  in  one  operation  if  you  can. 


I,  §8]  FUNDAMENTAL  NOTIONS  11 

15.  From  the  sum  of  2  s+8  £—4  w  and  3  s  — 6  t+2  w  take 
the  sum  of  8  s+9  Z+6  w  and  4  s-7  t-4  w. 

Remove  the  parentheses  and  combine  terms  in  each  of  the 
following  expressions. 

16.  a-(2a+4)-(5a+10). 

17.  6x+(5x-\2x+ll). 

[HINT.     Remove  the  innermost  group  'sign  first.] 

18.  x-\x-(x-3x)\. 

19.  2Qz-[(2z+7w)-(3z+5w)]. 

Find  the  values  of  the  following  when  o  =  4,  6  =  3,  c=  —  2, 
and  d=  —I. 

20.  10c2-(3a+6+d). 

[HINT.  Simplify  the  expression  as  far  as  possible  before  giving 
to  a,  6,  c,  and  d  their  special  values.] 

21.  3d-\a-(c-b)\. 

22.  o-Hc-(3d-6)j+3Ka-c)-7(&+<f)!. 

2c-(a2+b2) 
' 
24. 

8.  Multiplication  The  following  formulas  and  rules 
will  be  recalled  from  the  First  Course. 

Formula  I.  xmxn  =  xm+n. 

Thus          22-23=25;          x2-x3=x6;  (2  a)3  •  (2  a)4  =  (2  a)7; 

(a +6)5.  (a  +  &)8  =  (a+6)13;  etc. 

Formula  I  leads  to  the  following  rule. 

(a)  To  multiply  one  monomial  by  another,  multiply  the 
coefficients  for  the  new  .coefficient  and  multiply  the  letters  to- 
gether, observing  Formula  I. 

Thus,  in  multiplying  —4  w2n3  by  2  m2n2  the  new  coefficient  is 
(— 4)X2,  or  -8,  and  the  product  of  the  letters  is  m2n3m2n2,  or 
w2m2n3n2,  which  reduces  by  Formula  I  to  w4n5.  The  answer,  there- 
fore, is  —8  m*n6. 


12  SECOND  COURSE   IN  ALGEBRA  [I,  §  8 

Formula  II.  (xy)m  =  xmym. 

Thus        (2  •  3)2  =  22  •  32  ;       (xy)*  =  x*y*  ;       (3  y)3  =  33  •  y*  =  27  y*  ; 
(2  mn)2  =  (2  m  •  n)2  =  (2  m)2  •  n2  =22  •  m2  •  n2  =4  m2n2  ;   etc. 

Formula  III.  a(b-\-c)=ab-\-ac. 

Thus  2(3+4)=2-  3+2-  4=6+8  =  14;          x(y*+z*)  =xy*+xz3', 
ra2(mn2  —  m2n)  =  m3n2  —  m4n  ;  etc. 

Formula  III  leads  to  the  following  rules. 
(6)   To  multiply  a  polynomial  by  a  monomial  multiply  each 
term  of  the  polynomial  separately  and  combine  the  results. 

Thus  the  process  of  multiplying  the  polynomial  m—  n+mn  by 
the  monomial  mn  is  as  follows  : 

m  —n+mn 
_  mn 
m2n  —  mn2+m2n2.  Ans. 

(c)  To  multiply  one  polynomial  by  another,  multiply  the  mul- 
tiplicand by  each  term  of  the  multiplier  and  combine  results. 

Thus  the  process  of  multiplying  x—y+3z  by  2x+3y—z  is  as 
follows  : 

x-y+3z 


Multiplying  by  2  x,  2x*-2xy+6xz 
Multiplying  by  3  y,  3  xy  —  3  ?/2  +  9  yz 

Multiplying  by  —  z,  _  —    xz  _  +      yz—  3  z2 

Combining  results,  2  x-  +    xy  +5  xz  —3  y2  +  10  yz  —3  z2.     Ans. 

EXERCISES 

Find  the  product  in  each  of  the  following  indicated  multi- 
plications. 

1.  10a5x6a2.  4.   3  a262c3  X  -  2  aW. 

2.  -2a6x3a6.  5.   4  xyz2  X  -  8  x2yz. 

3.  -  4  ra2™3  X2  w2n.         6.    (  -  4  a26c2)  X  (2  a6)  X  (  -  3  ac)  . 
[HINT  TO  Ex.  6.    Multiply  the  first  two  expressions  together, 

then  multiply  this  product  by  the  last  expression.] 


I,  §  8]  FUNDAMENTAL  NOTIONS  13 

Simplify  each  of  the  following  expressions. 

7.    (3z)2.  8.    (3a6)2. 

[HINT  TO  Ex.  8.     See  fourth  illustration  under  Formula  II.] 

9.    (2  ran)3.  10.    (8  abc)2. 

[HINT  TO  Ex.  10.     8  abc  may  be  written  8  a  •  be.] 
11.    (-2wn)4.  12.    (-2z2?/2)3. 

13.  Show  that  if  the  side  of  one  square  is  twice  that  of 
another,  its  area  is  four  times  as  great. 

[HINT.     Let  a  =  a  side  of  the  small  square.     Then,  a  side  of  the 
large  square  =2  a,  and  the  area  =  (2  a)2.     Now  apply  Formula  II.] 

14.  Show  that  if  the  edge  of  one  cube  is  twice  that  of  an- 
other, the  volume  is  eight  times  as  great. 

15.  Compare  the  areas  of  two  circles,  one  of  which  has  a 
radius  three  times  as  great  as  the  other.    (See  Ex.  14  (b) ,  p.  6.) 

16.  Compare  the  volumes  of  two  spheres,  one  of  which  has 
a  radius  twice  as  great  as  the  other.     (See  Ex.  14  (e),  p.  6.) 

Find  the  product  in  each  of  the  following  multiplications. 

17.  (10a36+7a&4)X-2a26. 

18.  (2x2-3xy+5y2)x-2xy. 

19.  (a2- 10  a6+15  b2)  x4  a262. 

20.  (a— 2a6+62)x(a-6). 

21.  (2z+7)x(3z+5). 

22.  (4a2-106+l)x(2a2-6+2). 

23.  Simplify  the  expression  (2  x-\-y)2. 

SOLUTION.  '    (2  x  +y)2  =  (2x  +y)  •  (2  x  +y).      Multiplying  gives 
4xz+4xy  +  y2.     Ans. 

24.  If  the  side  of  a  square  is  represented  by  3  x  —  2,  what 
represents  its  area? 

[HINT.     Simplify  your  answer  as  in  Ex.  23.] 

25.  If  the  dimensions  of  a  rectangle  are  represented  by 
x+2  and  x—1,  what  represents  its  area? 


14 


SECOND  COURSE  IN  ALGEBRA 


[I,  §8 


26.   What  represents  the  area  of  the  triangle  whose  base 
is  2  x-\-3  and  whose  altitude  is  x  —  5? 

27.   What  represents  the  vol- 
ume of  the  sphere  whose  radius 


28.  Explain  how  the  figure  to 
the  left  illustrates  the  geometric 
meaning  of  Formula  III. 


FIG,  2. 


9.   Division.     The  following  formula  and  rules  will  be 
recalled  from  the  First  Course. 


Formula  IV. 


Thus        = 


(2o)4=24a4 


Formula  IV  leads  to  the  following  rules. 

(a)  To  divide  one  monomial  by  another,  divide  the  coeffi- 
cients for  the  new  coefficient,  observing  the  law  of  signs  for 
division   (§2    (e)),  and   divide  the  literal  factors,   observing 
Formula  IV. 

Thus,  in  dividing  28  a362  by  -4  ab  the  process  is  carried  out  as 
follows  :  _4  qb|  28  a3fr2 

—7  a?b.     Ans. 

Here  the  division  of  28  by  —4  gives  the  new  coefficient,  —  7, 
then  the  division  of  o3  by  a  gives  a2  (by  Formula  IV),  and  finally  the 
division  of  62  by  b  gives  b. 

(b)  To  divide  a  polynomial  by  a  monomial,  divide   each 
term  of  the  polynomial  separately,  and  combine  results. 

Thus  the  division  of  8  x2y—4  x3y*+2  xy  by  2  xy  is  carried  out 
as  below  :  2  xy  \  8  x*y-4  x*y*+2  xy 

4z     -2x2y  +1.      Ans. 


I,  §9]  FUNDAMENTAL  NOTIONS  15 

(c)    To  divide  one  polynomial  by  another  : 

1.  Arrange  the  dividend  and  divisor  in  the  descending  (or 
ascending)  powers  of  some  common  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  write  the  result  as  the  first  term  of  the  quotient. 

3.  Multiply  the  whole  divisor  by  the  first  term  of  the  quo- 
tient; write  the  product  under  the  dividend  and  subtract  it  from 
the  dividend. 

4.  Consider  the  remainder  as  a  new  dividend,  and  repeat 
steps  1,  2,  and  3,  continuing  in  the  same  manner  thereafter. 

Thus  the  division  of   17  x  +20+3  x2  by  4+z  is  carried  out  as 
follows  : 


3x2  +  l2x  3z+5     Quotient.     Ans. 

5z+20 
5s  +20 
0 

In  this  example,  a  new  dividend  is  finally  obtained  which  is  equal 
to  0.  Whenever  this  happens,  the  division  is  said  to  be  exact.  In 
cases  where  the  division  does  not  come  out  in  this  way,  there  is  a 
remainder,  as  illustrated  by  the  following  example  : 

3  x*  +  17  z+20  |     x+3 

3  x2  +  9  x  3  x  +8     Quotient.     Ans. 

8z+20 

8x+24 

—  4     Remainder. 


EXERCISES 

1.  How  do  you  test  (or  check)  to  see  whether  an  answer 
in  division  is  correct?  Is  the  same  method  used  for  this  in 
both  arithmetic  and  algebra?  Check  the  correctness  of  the 
results  obtained  for  the  last  two  examples  in  §  9. 

Perform  the  following  divisions,  and  check  your  answer. 


16  SECOND  COURSE   IN  ALGEBRA  [I,  §9 


2.  a*  +  a\  3.    (3  g)g+(3  g)\  4. 

5-  (iP2<z)7-(iP2<?)3.  8.   f7rr>-2  TTT. 

6.  -16zyz2-=-4z2/2z.          9.   3a6(a+6)2 

7.  4a462c3^20a262c.  10.    (9  m3np+18  wn3p)-=-3  mn. 

11.  (6  z2?/z+12  :n/2z-24  zi/z2)  -5-  (-3  xyz). 

In  each  of  the  following  divisions,  find  the  quotient,  also 
the  remainder  if  there  is  one.     Check  your  answer  for  each. 

12.  (3z2-2z-l)-=-(z^l).     13.  (15x2+x-2)-^-(3x-l). 
14.    (4</3+22/2-l)-K22/-l). 

[HINT.     Write  the  dividend  in  the  form  4  y3  +2  yz  +0  y  —  1.] 

15. 

16. 

17. 

18. 

19. 

20. 


MISCELLANEOUS   EXERCISES 

1.  Add  4x*-2x*-7x  +  l,  z3+3z2+5z-6,  4  x2-8 
2x3-2x2+8x+4,  and  -2  x  +  1+2  z3-3  x2. 

2.  Add  3(a+6)+6(6+c),  5(a+&) 

3(&+c)  -(a+6),  2(6+c)  -10(a  +  &),  and  3(a+6)  -3(6+c). 

3.  From  the  sum  of  1  +x  and  1  —  xz  subtract  1  —  x+x2  —  x3. 

4.  Simplify  the  expression 

ab  -{5  +x  -  (b  +c  -ab  +z)}  +  [x  -  (b  -  c  -7)]. 

6.    When  x=3,  m=6,  n=2  find  the  value  of  the  expression 

(m  +n  +x)n  —  (m  +n  —  x)n  —  (m  -  n  +z)"(  -  m  +n  +x)n. 

6.  Simplify  the  expression 

y3-[2x3-xy(x-y)-y*]+2(x-y)(x*+xy+y*). 

7.  Find  the  remainder  in  the  following  divisions 

(a)   (0^  +  1)^(0-1). 

(6)     [X3«-3  +2/3n+3]  ^.  [xn-l 


CHAPTER  II 
SPECIAL   PRODUCTS   AND   FACTORING 

10.  Special  Products.  Certain  products  occur  so  fre- 
quently in  algebra  that  it  is  desirable  to  study  them  with 
especial  care  and  to  remember  their  forms.  In  this  connec- 
tion, the  following  formulas  will  be  recalled. 

Formula  V.          (x-y)(x+y  =  x2-yz. 

Thus  (x-8)(x+8)=x2-82=a;2-64; 
(9  x-3  y)(9  x+3  y)  =(9  z)2-(3  yY  =81  x2-9  y\ 


Formula  VI.        (x+yY~  =  x2+2  xy+y2. 

Thus  (r+6)2  =  r2+2(r-  6)  +62  =r2  +  12  r+36; 


Formula  VII.        (x-y)z  =  x2-2  xy+y2. 

Thus  (r-6)2=r2-2(r-  6)  +62  =r2-12  r+36; 


Formula  VIII.     (x+m)(x+n)  =  x2  +  (m+n)x+mn, 
m  and  n  being  any  (positive  or  negative)  numbers. 


Thus  (x+4)(x+3)=x2  +  (4+3)x+4-  3=z2+7z+12; 

(-6 
17 


18 


SECOND   COURSE   IN  ALGEBRA 


[II,  §  10 


ORAL  EXERCISES 

State  under  which  formula  each  of  the  following  products 
comes,  and  read  off  the  answer  by  inspection. 

1.  (z-3)(z+3).       3.  (6+a)(6-o).      5.  (5  x-2)(5z+2). 

2.  (r-5)(r+5).       4.  (x-4)(x+4).      6.  (1- 

7.  (xy-12)(xy+12). 


8'  (HXH 


9.  (z+8)2. 

10.  (z-8)2. 

11.  (2z+l)2. 

12.  (3z-4)2. 

13.  (a+26)2. 
24.    4 


14. 
15. 
16. 

17.  (3x-f)2. 

18.  (2 

2).  26. 

25.   (3o-5|/)(3o+7y).     27. 

28.    }8-(r+«)H8+(r+«)}. 

29. 

30. 


19. 

20.  (a+6)(a-8). 

21.  (a;- 
22. 

23.  (a2+4)(a2-6). 


WRITTEN   EXERCISES 


1.  Show  how  Formulas  VI  and  VII  can  be  obtained  as 
special  cases  of  Formula  VIII. 

2.  Show  how  the  following   three   figures   illustrate   re- 
spectively the  geometric  meanings  of  Formulas  V,  VI,  and  VII. 


-y— 


(x-y)2 


II,  §11]       SPECIAL  PRODUCTS  AND  FACTORING  19 

By  use  of  Formulas  V,  VI,  and  VII  write  down  (without 
multiplying  out)  the  simplest  forms  for  the  following  prod- 
ucts. 

3.  (a2+a-l)(a2-a+l). 

[HINT.  Write  first  as  {a2  +  (a-l)}{a2-(a-l)},  then  apply  For- 
mula V,  afterwards  simplifying  your  answer  by  Formula  VII.  The 
final  result  is  a4  -a2  +2  a  -  1.] 

4.  (a2+a6+62)(a2-a&+&2).  7.    J(z+7/)-4(2. 
6.    (x2+x-2)(x*-x-2).                 8.    [7+(m-n)]2. 

6.    (a-b+m+n)(a-b-m-n).     9.    [(x+y)-(m+ri)]2. 
•  10.    (a+6+c)2. 

[HINT.     Write  as  [(a  +6)  -fc]2  and  apply  Formula  VI  twice.] 
11.   The  rule  for  finding  the  square  of  any  polynomial  is 
as  follows.     The  square  of  any  polynomial  is  equal  to  the  sum 
of  the  squares  of  its  terms  plus  twice  the  product  of  each  term  by 
each  term  that  follows  it.     For  example, 


a2+62+c2+d2+2  ab+2  ac+2  ad+2  bc+2  bd+2  cd. 

Show  that  Formulas  VI  and  VII  conform  to  this  general 
rule  ;  also  that  your  answer  for  Ex.  10  does  so. 

By  means  of  the  general  rule  in  Ex.  11,  write  out  the  values 
of  each  of  the  following  expressions. 

12.  (a+6-c)2.  14.    (2x+y-z)2. 

13.  (a-b-c)2.  15.    (2x+2y-z+3w)2. 

11.  Type  Forms  of  Factoring.  Factoring  is  the  reverse 
of  multiplication  in  the  sense  that  in  multiplication  certain 
factors  are  given  and  we  are  asked  to  find  their  product, 
while  in  factoring  a  certain  product  is  given  and  we  are 
asked  to  find  its  factors,  that  is,  to  find  expressions  which 
multiplied  together  produce  it.  The  following  four  types 
of  expressions  are  to  be  especially  noted,  as  they  can  always 
be  readily  factored. 


20  SECOND  COURSE  IN  ALGEBRA  [II,  §  11 

(a)  Expressions  whose  terms  each  contains  a  common  factor. 
Thus      mx+my+mz  =  m(x+y+z);   (Formula  III,  p.  12.) 

a?x+ax*+a2xz=ax(a+x+ax)  ; 

ax  —  ay+bx-by=a(x-y)  +b(x  —  y)  =  (x  —  y)(a+b)  ; 
2  x  -y  +4  x2  -2  xy  =  (2  x  -  y)  +2  x(2  x  -  y}  =  (2  x  -  y)  (1  +2  x). 

Note  that  in  all  these  examples  the  given  expression  has 
finally  been  exhibited  as  a  product.  This  is  essential  to  every 
example  in  factoring. 

(b)  Expressions  which  can  be  regarded  as  the  difference  of 
two  squares. 

Thus      25x*-yz  =  (5x-y)(5x+y)',  (Formula  V,  p.  17.) 
a262  -c2d2  =  (ab  -cd)(ab  +cd)  ; 


(c)  Trinomials  of  the  form  x2  -\-px-\-q,  where  p  and  q  have 
such  values  that  we  can  readily  find  two  numbers  whose  sum  is 
p  and  whose  product  is  q. 

Thus,  in  factoring  z2+7  a;  +  12,  we  need  only  inquire  whether  we 
can  find  two  numbers  whose  sum  is  7  and  whose  product  is  12. 
The  numbers  3  and  4  are  seen  (by  inspection)  to  do  this.  Hence  we 
know  by  Formula  VIII  that  we  may  write 

z2+7z+12  =  (z+3)(z+4).     Ans. 
Similarly,  z2-z-12  =  (z-4)(z+3)  ;  Why? 

z2-5  xy-36  y*  =  (x-9  7/)(z+4  y)  ; 
a2&2  -21  ab  -72  =  (ab  -24)  (ab  +3). 

(d)  Trinomials  of  the  form  axz-\-bx+c  which  are  perfect 
squares,  that  is  such  that  the  coefficients  a  and  c  are  perfect 
squares  while  the  other  coefficient,  namely  6,  is  equal  in  ab- 
solute value  (§  1)  to  twice  the  product  of  the  square  roots  of 
a  and  c. 

Thus  9  z2  +  12  z+4  is  a  perfect  square  because  12  =2  •  A/9  •  V4. 
Hence,  by  Formula  VI,  we  have 

9z2  +  12z+4  =  (3z+2)2  =  (3z+2)(3:r+2).     Ans. 
Similarly,  xz  —  l4x  +49  is  a  perfect  square,  because 
14=2-  VI- 


II,  §  11]      SPECIAL  PRODUCTS  AND  FACTORING  21 

Thus  we  have,  using  Formula  VII, 

x2  -  14  x  +49  =  (x  -7)2  =  (x  -7)  (x  -7). 
For  a  like  reason,  we  see  that  z2  +  14  x+49  =  (x+7)(x+7). 

The  following  are  two  other  examples  that  can  be  brought 
under  this  case. 

a2b2  -2  ab  +4  =  (ab  -2)2  =  (ab  -2)  (ab  -2). 


We  could  not,  however,  factor 
by  this  method.     Why  ? 


EXERCISES 

Factor  each  of  the  following  expressions  : 
1.    (a)  .z2+2  x.  (c)   8  a2+24  a. 

(6)  x*y+xy*.  (d) 

(e) 


—  c)—  g(a+&—  c). 
W  pq  —  px  —  rq+rx. 
(i)   y*-±y+xy-±x.     (j)   3  x*-15x+Wy-2x*y. 


2.    (a)  81  -x2.  (/)   962-(a-x)2. 

(6)  a2-62c2.  (g)  49  a2-(5  a-4  6)2. 

(c)  144z2-4.  (h)  (2x+5)2-(5x-3) 

(d)  ixV-36.  (i)    (x+x2)2-(2x+2)2. 

W       ~-  0) 


3.    (a) 

(6)  x2-6o:+8.  (g)   12+7  a+a2. 

(c) 


(e)  x2-o;-110.  0') 


22 


SECOND  COURSE   IN  ALGEBRA 


[II,  §  11 


16-24(a-6)+9(a-6)2. 


4.   Test  each  of  the  following  expressions  to  see  whether 
it  is  a  trinomial  square,  and  if  so,  factor  it. 

(a)  z2-8z+16.  (/) 

(6)  z2-12z+36.  fa) 

(c)  4z2+6z+l.  (h) 

(d)  81z2+18z+l.  (i) 

(e)  81-72r+16r2.  (f) 

6.   The  figure  shows  a  square  of  side  a  within  which  lies 
(in  any  manner)  a  smaller  square  of  side  b.     Prove  that  the 
area  between  the  two  (shaded  in  the  figure) 
is  equal  to  (a +&) (a  —  b). 

6.  The  result  in  Ex.  5  furnishes  a  rule 
for  determining  quickly  the  area  between 
any  two  squares  when  the  one  lies  within 
the  other.     State  the  rule. 

7.  By  means  of  Exs.  5  and  6  answer  the 
following :  What  is  the  area  of  pavement 

in  the  street  surrounding  a  city  block  one  half  mile  on  a 
side,  the  street  being  4  rods  wide?     (1  mile  =  320  rods.) 

8.  Show  that  the  area  between  a  circle  of  radius   R 
and  a  smaller  circle  of  radius  r  lying  within  it  is  equal  to 
ir(R+r)(R— r).      Does  it  make   any  difference  where  the 
smaller  circle  lies  so  long  as  it  is  within  the  large  one? 

9.  Show  that  if  a  and  b  are  the  sides  of  a  right  triangle 
whose  hypotenuse  is  h,  we  shall  have  a2=  (h+b)(h  —  b) ;  also 
V=(h+a)(h-a). 

10.  Formula  V  is  frequently  used  to  find  the  square  of  a 
number  quickly  by  mental  arithmetic.  Suppose,  for  example, 
that  we  wish  to  know  the  value  of  162.  We  first  take  6  away 
from  the  number,  leaving  10,  then  we  add  6  to  it,  giving  22. 


II,  §  12]       SPECIAL  PRODUCTS  AND  FACTORING          23 

We  multiply  the  10  and  the  22  thus  obtained  (as  is  easily 
done  mentally) ,  giving  220.  Now  all  we  have  to  do  is  to  add 
62,  or  36,  to  the  220  to  obtain  the  desired  value  of  162,  giving 
256  as  the  answer.  The  reason  for  these  steps  appears  below. 

(16-6)(16+6)  =  162-62,     (Formula  V.) 
whence     (16-6)(16+6)+62  =  162. 

Find  (mentally)  in  this  way  the  value  of  each  of  the  fol- 
lowing expressions. 

(a)  152.  (c)   172.  (e)  312. 

[HINT.     Subtract  5  first,  then  add  5.] 

(6)  142.  (d)  22*.  (/)  452. 

*  12.  Other  Type  Forms.  Besides  the  type  forms  men- 
tioned in  §  11,  the  following  may  be  noted  : 

(e)  Trinomials  of  the  form  ax^+bx+c  which  are  not  per- 
fect squares  and  hence  do  not  fall  under  (d)  of  §  11.  There  is 
no  general  rule  in  such  cases,  though  we  may  frequently 
discover  by  inspection  whether  a  given  trinomial  of  this  form 
is  factorable  readily,  and  if  so,  obtain  its  factors.  This  is 
best  understood  from  an  example. 

EXAMPLE  .    Factor  15  x2— 7  x  —  2. 

SOLUTION.  For  15  x2,  try  5  x  and  3  x  ;  thus  we  begin  by  writing 
(5  x  ) (3  x  ),  where  the  open  spaces  are  yet  to  be  filled  in. 

For  —2  try  1  and  2  with  unlike  signs,  arranging  the  signs  so  that 
the  sum  of  the  cross  products  shall  give,  as  desired,  the  —  7  x  of  the 
given  expression;  thus  we  now  try  (5  z  +  l)(3  x—  2).  Here  the 
middle  term  of  the  product  (cross  product)  is  readily  found  to  be 
—  10  x+3  x,  or  —  7  x,  as  desired.  The  only  other  possibility  would 
be  (5  x  — 1)(3  x+2),  but  as  the  cross  product  term  here  becomes 
10  x  —3  x,  or  +7  x,  this  form  cannot  be  the  one  we  desire. 

We  have,  therefore,  15  x2  -7  x  -2  =  (5  z  +  l)(3  x  -2).     Ans. 


24  SECOND  COURSE   IN  ALGEBRA  [II,  §  12 

(/)   The  sum  of  two  cubes.     This  form  is  factorable  in  ac- 
cordance with  the  following  formula. 

Formula  IX.     x*+y*=(x+y)(x*-xy+y*). 

Thus  x3  +8  =  x3  +23  =  (x  +2)  (x2  -2  x  +22) 

=  (x+2)(x2-2x+4).    Ans. 
Likewise, 

W3n3  +64  p3  =  (ran)3  +  (4  p)3  =  (ran  +4  p)  [(ran)2  -  (ran)  (4  p)  +  (4  p)2] 

=  (ran+4  p)(ra2n2—  4  ranp  +  16  p2). 


(g)  The  difference  of  two  cubes.     This  form  is  factorable 
in  accordance  with  the  following  formula. 

Formula  X.     x*-y*=(x-y)(x2+xy+y*). 


-27=x3-33  =  (x-3)(z2+3x-f32) 

=  (x-3)(x+3x+9).    Ans. 

8  x3  -  125  2/3  =  (2  x)3  -  (5  y)*  =  (2  x  -5  y)[(2  x)2  +  (2  x)  (5  y)  +  (5  y)2] 

=  (2  x-5  2/)(4  x2  +  10  xy+25  i/2).     Ans. 


13.  Complete  Factoring.  Each  of  the  exercises  on  p.  21 
concerns  but  a  single  one  of  the  type  forms  mentioned  in  §  11, 
but  we  often  meet  with  problems  in  which  two  or  more  of  the 
types  are  concerned  at  the  same  time.  Thus,  in  factoring 
as£  _  a^}  we  first  take  oirt  the  common  factor  ab.  This  gives 
a?b  —  abs  =  ab(a2-b*).  But  a2  —  62  is  itself  factorable,  coming 
under  type  (6)  of  §  11.  The  final  answer,  therefore,  is 
ab(a-b)(a+b). 

Other  illustrations  of  this  idea  occur  below.  Note  that  the 
final  answer  in  every  case  contains  no  factors  which  them- 
selves can  be  still  further  broken  up  into  other  factors. 

EXAMPLE  1.    Factor  completely  x*-y2+x-y. 
SOLUTION. 


,  §11.) 
Ans. 
(See  (a)   §11.) 


11,  §  13]       SPECIAL  PRODUCTS  AND  FACTORING  25 

EXAMPLE  2.     Factor  completely  a4 +3  a2b2— 4  64. 

SOLUTION. 

o4+3  a262-4  64  =  (a2+4  62)(a2-62)  (See  (c),  §  11.) 

=  (a2+4&2)(a-6)(a+6).     (See  (6),  §  11.) 

EXERCISES 

Factor  completely  each  of  the  following  expressions.     Those 
preceded  by  the  *  involve  the  type  forms  mentioned  in  §  12. 

1.  x*-x2-x+l.  14. 

[HINT.     Write    in     the    form.       15. 
z2(z-l) -(*-!).]  *16. 

2.  2x*-8x2y+Sxy2.  *17. 

[HINT.     Write     in     the     form       18- 
4xi/-[-4i/2).]  *19. 

3.  x2+3ax-3a-x.  *20.    20z2-6z-2. 

4.  a3+2a2-4a-8.  21-   Sx*~ 

5.  x4-lQ^2J-Qft  22>   s4-1* 


^   i    . .4.       r»  —9.. 9  ^O. 

. 


24. 

^JLuLOflfc  25t   9^4-30^2a+25a2. 

26. 


[HINT.       1  —  a2  —  62  +  2  a6  =  1  —  o  i.- 

(a-6)2.]  ~ 

27.    (a-x2)2-(6-i/2)2. 

9.   m2-4mn+4n2-16.  *28 


10.  2xy-x  -y2+l.  29.  x2-9  t/2+a:+3  y. 

11.  l+9c2+6c.  30.  a3-2a26+4a62-863. 

1O  Q  /y»3  ^_  Q  /Y»    I    /I   /y»4       /|   /yi2  QO  />»4 /1 7  /v*2 

•Lv«  O  •*/  O  »i/    \     A  «|/  A  «v  •  O*i»  »i/    """"  A  I    •*/ 


CHAPTER  III 

HIGHEST  COMMON  FACTOR  AND  LOWEST  COMMON 

MULTIPLE 

14. .  Prime  Factor.  A  number  which  has  no  factor  except 
itself  and  unity  is  called  in  arithmetic  a  prime  number. 
Such  a  number  when  used  as  a  factor  is  called  a  prime  factor. 

Thus  the  prime  factors  of  15  are  3  and  5. 

The  word  prime  factor  is  similarly  used  in  algebra. 
Thus  we  say  that  the  prime  factors  of  3  abc  are  3,  a,  6,  and  c. 
The  prime  factors  of  18  x2y  are  2,  3,  3,  x,  x,  and  y. 
The  prime  factors  of  o26(a2  —  b2)  are  a,  a,  b,  a  —  6,  and  a  +b.      (See 
Formula  V.) 

15.  Finding  Common  Factors.     As  soon  as  we  have  fac- 
tored each  of  several  expressions  into  its  prime  factors,  we  can 
readily  pick  out  their  common  factors  (§5). 

Thus,  in  finding  the  common  factors  of  abc,  a?b,  a&2,  and  3  ab,  we 
write 
abc  =  a  •  b  •  c,         azb  =  a  •  a  •  b,         ab2  =  a  •  b  •  b,         3  ab  =  3  •  a  •  b. 

The  common  factors  are,  therefore,  a  and  6,  since  these  occur  in 
each  expression  and  they  are  the  only  factors  thus  appearing. 

16.  Highest  Common  Factor.     The  product  of  all  the 
common  prime  factors  of  two  or  more  expressions  is  called 
their  highest  common  factor.     It  is  called  the  highest  be- 
cause it  contains  all  the  common  factors,  and  the  usual  ab- 
breviation for  it  is  H.  C.  F. 

EXAMPLE  1.    Find  the  H.  C.  F.  of  lOz2!/3,  2xy2,  and  18sV- 
SOLUTION.     Resolving  each  into  its  prime  factors, 
10zV=2-  5-  x-  x-  y  y  y, 

2xy*=2-  x-  y  y, 
18zV  =2-  3'3-x-X'X-yy. 

The  factors  common  to  the  three  expressions  are  thus  seen  to  be 
2,  x,  y,  y. 

The  H.  C.  F.  is,  therefore,  2  •  x  -  y  -  y,  or  2  xy*.     Ans. 

26 


Ill,  §  17]    COMMON  FACTORS  AND  MULTIPLES  .  27 

EXAMPLE  2.  Find  the  H.  C.  F.  of  3  z2+3  x-18, 
6  z2+36  z+54,  and  9  z2-81. 

SOLUTION. 

3z2+3z-18=3(z2-fz-6)=30c+3)(z-2).  (See  (c),  §11.) 

6z2+36z+54=6(z2+6-z+9)=2.  3(z+3)(z+3).      (See  (d),  §11.) 

9z2-81=9(z2-9)=3-  3(x+3)(x-3).  (See  (6),  §11.) 

The  common  factors  being  3  and  (x  +3),  the  H.  C.  F.  is  3(x  +3.)   Ans. 

In  general,  to  find  the  H.  C.  F.  of  two  or  more  expressions  : 

1.  Find  the  prime  factors  of  each  expression. 

2.  Pick  out  the  different  prime  factors  and  give  to  each  the 
lowest  exponent  to  which  it  occurs  in  any  of  the  expressions. 

3.  Form  the  product  of  all  the  factors  found  in  step  2. 

NOTE.  Since  the  H.  C.  F.  of  several  expressions  consists  only 
of  factors  common  to  them  all,  it  is  always  an  exact  divisor  of  each 
of  the  expressions.  It  is  therefore  called  in  arithmetic  "the  great- 
est common  divisor"  and  is  represented  by  G.  C.  D. 

EXERCISES 

Find  the  H.  C.  F.  of  each  of  the  following  groups. 

1.  12,  18.  7.  a2+7a+12,  a2-9. 

2.  16,24,36.  8.  x2-y2,(x-y)2,x2-Zxy+2y*. 

3.  x2y,  xy2.  9.  ra2+4  m+4,  m2-6  m-16. 

4.  a2b,  ab2,  a2b2.  10.  3  ?/2-363,  y2-7  T/-44. 

6.  2  x*y,  6  x*y,  14  x*y2z4.     11.  2  a2+4  a,  4  a3+12  a2+8  a. 

6.  a2-62,  a2-2a&+62.       12.     4-i/4,  x*+xzy+xy2+y*. 

13.  3  r5+9  r4-3  r3,  5  rV+15  rs2-5  s2,  7  ar2+21  ar-7  a. 

*14.  a3-63,  a2-62,  a-b.      *15.  x3+l,  x2-x+l. 

17.  Common  Multiple.  In  arithmetic  a  number  which 
is  exactly  divisible  by  two  or  more  given  numbers  is  called  a 
common  multiple  of  them. 

The  word  common  multiple  is  similarly  used  in  algebra. 

Thus  4  x2y2  is  a  common  multiple  of  x  and  y ;  it  is  also  a  common 
multiple  of  4  and  x.  Similarly,  a2  —  b2  is  a  common  multiple  of  a  —  b 
and  a +6. 


28  SECOND  COURSE   IN  ALGEBRA          [III,  §  18 

18.  Lowest  Common  Multiple.  The  lowest  common 
multiple  of  two  or  more  numbers  or  expressions  is  that 
multiple  of  them  which  contains  the  fewest  possible  prime 
factors.  Its  abbreviation  is  L.  C.  M. 

The  following  examples  illustrate  what  the  L.  C.  M.  means 
and  how  to  obtain  it. 

EXAMPLE  1.  Find  the  L.  C.  M.  of  10  a2b,  16  a263,  and 
20  a364. 

SOLUTION.    Separate  each  expression  into  its  prime  factors.   Thus 
10a26=2-  5-  a-  a-  6, 
16a263=2-  2-  2-2-  a-  a-  6-  6-  6, 
20a364=2-  2-5-a-a-a.6-6.b-6. 

The  L.  C.  M.  is  2  -  2  -  2  •  2  •  5  •  a  •  a  -  a  •  6  •  6  -  6  -  6,  or  80  a364, 
since  this  contains  all  the  factors  of  each  of  the  three  expressions, 
and  at  the  same  time  is  made  up  of  fewer  factors  than  any  other 
similar  expression  that  can  be  found. 

EXAMPLE  2.     Find    the    L.  C.  M.    of    z4-10z2+9    and 


SOLUTION. 


Therefore  the  L.  C.  M.  is  (z- 

In  general,  to  find  the  L.  C.  M.  of  two  or  more  expressions  : 

1.  Find  the  prime  factors  of  each  expression. 

2.  Pick  out  the  different  prime  factors,  taking  each  the  great- 
est number  of  times  it  occurs  in  any  one  of  the  expressions. 

3.  Form  the  product  of  all  the  factors  found  in  step  2. 
NOTE.     From  the  manner  in  which  the  L.  C.  M.  is  formed,  it 

must  be  exactly  divisible  by  each  of  the  given  numbers,  or  expressions. 

EXERCISES 

Find  the  L.  C.  M.  of  each  of  the  groups  of  expressions  in 
the  exercises  on  p.  27. 


CHAPTER  IV 
FRACTIONS 

19.  Definitions.     Any  expression  of  the  form  a/6  is  called 
a  fraction.    It  means  the  number,  or  expression,  which  when 
multiplied  by  b  gives  a.    The  part  above  the  line,  or  a,  is 
called  the  numerator,  while  the  part  below  the  line,  or  6,  is 
called  the  denominator.    The  numerator  and  denominator 
taken  together  are  called  the  terms  of  the  fraction. 

20.  Equivalent  Fractions.     It  is  often  desirable  to  change 
the  form  of  a  fraction  without  changing  its  value.     Such 
changes  all  depend  upon  the  following  principle. 

The  numerator  and  denominator  of  a  fraction  may  be  multi- 
plied or  divided  by  the  same  number,  or  expression,  without 
changing  the  value  of  the  fraction.  Thus 

3^3-  2^6. 
4    4-2     8' 
4  a_4  a  •  a_4  a2. 
5        5  •  a       5  a' 
a+b 
(a+6)2 


21.  Changes  of  Signs  in  Fractions.  There  are  three  signs 
to  be  considered  in  a  fraction  ;  the  sign  of  the  numerator,  the 
sign  of  the  denominator,  and  the  sign  of  the  fraction  itself. 

_  o 

Thus  in  H  --  ,  the  three  signs  in  the  order  just  mentioned  are 
4 

-,  +,  +,  while  in  --^7  they  are  +,  -,  -. 

—  D  0 

29 


30  SECOND  COURSE   IN  ALGEBRA  [IV,  §  21 

Since  a  fraction  is  merely  an  indicated  division,  the  law 
of  signs  for  division  (§  2  (e),  p.  3)  must  hold  at  all  times,  so 
that  we  arrive  at  the  following  rule. 

Any  two  of  the  three  signs  of  a  fraction  may  be  changed  with- 
out altering  the  value  of  the  fraction.  Thus 

_i_±3  =  +:d*  =      -3=     ±3 
+4         -4         +4         -4* 
Likewise 

a  _  — a  _      —  a  _        a 
b     -b       ~  5  _&' 

Care  must  be  taken,  however,  in  changing  the  sign  of  the 
numerator  or  denominator  of  a  fraction  when  polynomials 
are  present.  For  example,  if  the  numerator  is  a  polynomial, 
we  can  change  the  sign  of  the  whole  numerator  only  by  chang- 
ing the  sign  of  every  term  in  it.  A  similar  statement  applies 
when  the  denominator  is  a  polynomial.  Thus, 

Q+26+c     _       —a— 2b—c^  a +2  b+c 

2a-3b-2c        2a-3b-2c         -2o+36+2c* 

Observe  carefully  the  reason  for  every  change  of  sign  here. 


22.  Reduction  of  Fractions  to  Lowest  Terms.  A  fraction 
is  reduced  to  its  lowest  terms  when  its  numerator  and  de- 
nominator have  no  common  factor  except  1. 

To  reduce  a  fraction  to  its  lowest  terms,  factor  numerator  and 
denominator,  then  divide  each  by  all  their  common  factors. 


E 


EXAMPLE  2.    a2  -Ha  +24  =  (a- 
a2-a+6        (a- 


IV,  §23]  FRACTIONS  31 

23.  Lowest  Common  Denominator.  The  lowest  common 
denominator  of  two  or  more  fractions  is  the  lowest  common 
multiple  (§  18)  of  their  denominators. 

To  reduce  several  fractions  to  their  lowest  common  denomi- 
nator : 

1.  Find  the  L.  C.  M.  of  the  denominators. 

2.  For  each  fraction,  divide  this  L.  C.  M.  by  the  given  de- 
nominator, and  multiply  both  numerator  and  denominator  by  the 
quotient. 

EXAMPLE.  Reduce  the  following  fractions  to  equivalent 
fractions  having  their  lowest  common  denominator  : 

and 


SOLUTION.     Factoring  the  denominators,  the  fractions  may  be 
written 


The  L.  C.  M.  of  these  denominators  is  (a;  +3)  (a;  -3)  (z  +6).  In 
order  to  give  the  first  fraction  this  L.  C.  M.  as  its  denominator, 
multiply  its  numerator  and  denominator  by  x+G  (this  being  the 
L.  C.  M.  divided  by  the  denominator  of  the  first  fraction). 

In  order  to  give  the  second  fraction  this  L.  C.  M.  as  its  denomina- 
tor, multiply  its  numerator  and  denominator  by  x—  3  (this  being 
the  L.  C.  M.  divided  by  the  denominator  of  the  second  fraction). 
The  desired  forms  are,  therefore, 

(x+2)(x+Q)  d 


(x  +3)  (x  -3)  (x  +6)  (x  +3)  (x  -3)  (x  +6) 

Observe  that  these  fractions  are  respectively  equivalent  (  §  20)  to 
those  with  which  we  started,  but  these  have  denominators  that  are 
alike,  which  was  not  the  case  with  the  original  forms. 


32  SECOND  COURSE  IN  ALGEBRA  [IV,  §  23 


EXERCISES 

Write  each  of  the  following  fractions  in  three  other  ways 
without  changing  the  value. 

1          5  6  3    a-b_         .  3a-4 

-6*         '   l-x  '  c-d  '   (2z-l)i 


Reduce  each  of  the  following  expressions  to  lowest  terms. 
240 

'          *  ' 


320  (a6+l)2 

10  xy  4z2-2/2 

O.  *  11.      — • 

OA    /y>2/)/2  ni  O    /y. 

o\j  ju  y  y  —  £i  jc 

„     36  xr3  19    s2-6s+8 

*•    ^r — r~r*  J-^. 


72^^  s2-5s+6 

-9a252c2  1Q     r4-6r2+5 

o.    -3-: — r77~r~*  !*• 


54  aW  r2-6r+5 


28xy2-12x2y 

Reduce  all  of  the  fractions  in  each  of  the  following  groups 
to  the  lowest  common  denominator. 


16  2f  ^  £.  go     2  -^  __  L, 
'  4'6'2  '    62'a2-62'  a+b 

17  5  ?.  21 

'  a  6  ' 


a        r 


y*>x-y  '    (a+6)2'  l'a2-62 

1  1 


23. 


0.          a;-2  rc-1  o:+3 

Z4. 


x2-2x-8'  x2-3x-10'a;2- 


IV,  §  24]  FRACTIONS  33 

24.  Addition  and  Subtraction  of  Fractions.  The  follow- 
ing rule,  which  is  the  same  in  algebra  as  in  arithmetic,  will  be 
recalled  from  the  First  Course,  p.  151. 

To  add,  or  subtract,  fractions  : 

1.  Reduce  the  fractions  to  equivalent  fractions  having  their 
lowest  common  denominator  (L.  C.  D.). 

2.  Add,  or  subtract,  each  numerator  according  to  the  sign 
before  the  fraction  and  write  the  result  over  the  L.  C.  D. 

3.  Reduce  the  resulting  fraction  to  its  lowest  terms. 

EXAMPLE.     Simplify 


a-1     a+1     a2-! 

SOLUTION.     — Qn2+_3_a2_=4  a+4_a2-3  a+2+^g2_ 

a-1     a  +  1     a2-!      a2-!  a2-!          a2-! 

_4a+4-(a2-3a+2)+3a2_2a2+7a-+2 
a2-!  a2-! 

EXERCISES 

Simplify  each  of  the  following  expressions. 
~3~    ~4~    2'  x       y  *'  abz    ab     a? 

2-2f+£?"     4-^-2J1^+I-    6-H+r 


7  _ | _..  12 


2x-2    3(z+l)     3z-3  a2+2a6+62    a?-2ab+b2 

l3- 


9  _          .  14  _  . 


a—b    a+b  n+4     1-n    n2+3  n-4 


10.  ___  -+-~     15.  x-3+ 

r2-2r-8    r-4    r+2  x2+3  z+9 

11  1  1  16  _x__       1+a^2 

'  x-l     x2+x+l 


34  SECOND   COURSE   IN  ALGEBRA          [IV, 

25.  Multiplication  and  Division  of  Fractions.  Fractions 
are  multiplied  in  algebra  as  in  arithmetic  by  taking  the 
product  of  the  numerators  for  the  numerator,  and  the  product 
of  the  denominators  for  the  denominator,  canceling  wherever 
possible. 

EXAMPLE  1. 


4  xy*     9  ab3      4 

Canceling  like  factors  from  numerator  and  denominator,  this 
reduces  to 

^-     Ans. 
662 

EXAMPLE  2. 

a2  +2  a  -8  _  a2+5  a-6 
a2  +6  a         a2+a-12 


.a(a-3) 

In  algebra  we  divide  one  fraction  by  another  as  we  do  in 
arithmetic,  by  inverting  the  divisor  and  proceeding  as  in 
multiplication. 

EXAMPLE. 

a2-b2        .    a-b  a2-b2 


a-b 

A 


EXERCISES 


Perform  each  of  the  following  multiplications. 
3.,  2 


5  rs2      1*8     m3n3 

7  o         1^2  O 

o     «    5.  A    a      k         2 

1    6    c  '  a2+&2 

2/frl  /d_   O'1^  P\   /• 

CM/  *±  •*/  M  «J  « 


3  xy    5  a2  2  z2+4  Z7/+2  i/2       5 

.       6  a  \    10  b8  8     r+2       g-1      4s-6 

1  '  8  a3'  '  2s-3  *2r+2*   r+2" 


IV,  §  25]  FRACTIONS                                         35 

9. 

10  .        .  -  ___  -. 

x  —  ?/  6  —  a     £+?/    a—  6 
Perform  the  following  divisions. 

4^2 
'       * 


9*3  a(a+6)  '2  0-6 

12     10  &  .  12  s  r2-9s2 

*          '  15< 


13. 


11  y       y2  r2-4s2     r2-rs-6 

a2+2a,  (a+2)2 
a2-2a*  (a-2)2' 


Perform  the  indicated  operations  and  simplify  each  of  the 
following  expressions. 


16 

\x+y    x-yj     \x+y    x—yj 


is 

' 


3,5 

Li.       19. 

3_4  V   m       m-3\m-2 

x     y 


21.  fi- 


1  1  1 


22  . 


-X       1+05 


CHAPTER  V 
SIMPLE  EQUATIONS 

26.  Preliminary  Considerations.  Suppose  we  wish  to 
divide  64  into  two  parts  such  that  if  one  part  be  divided  by 
5  and  the  other  by  7  the  sum  of  the  quotients  shall  be  10. 
Such  a  problem  as  this  can  be  done  only  with  some  difficulty 
by  arithmetic,  but  it  is  a  simple  task  by  algebra. 

SOLUTION.     Let  x  represent  one  part. 

Then  64—  x  will  be  the  value  of  the  other  part. 

From  the  statement  of  the  problem,  we  are  to  have 


Let  us  multiply  both  sides  of  this  equality  by  35  (as  we  may  do 
without  destroying  it),  thus  clearing  it  of  fractions.     This  gives 
7  x  +320  -5  x  =350. 

Subtracting  320  from  both  sides  of  this  last  equality  (as  we  may 
do  without  destroying  it),  and  replacing  7  x  —5  x  by  its  value  2  x, 
we  obtain  2  x=  30. 

Hence  (dividing  both  sides  by  2)  we  have 

z  =  15. 

The  two  parts  sought  are  therefore  15  and  64  —  15,  or  49.     Ans. 

CHECK.  +      =  3  +7  =  1°- 


A  statement  of  equality,  like  any  of  those  above,  wherein 
a  single  unknown  letter  occurs,  and  occurs  to  no  higher  power 
than  the  first,  is  called  a  simple  equation.  It  is  also  known 
as  a  linear  equation,  or  an  equation  of  the  first  degree. 

The  process  of  finding  the  value  of  the  unknown  letter  is 
called  solving  the  equation. 

The  value  of  the  unknown  letter  is  called  the  solution, 
or  root,  of  the  equation. 


V,  §  27]  SIMPLE   EQUATIONS  37 

27.  Principles  Useful  in  the  Solving  of  Equations.  In 
solving  an  equation  we  may  at  any  point  in  the  process  add 
the  same  amount  to  both  sides,  or  subtract  the  same  amount 
from  both,  or  we  may  multiply  both  by  the  same  amount,  or 
divide  both  by  the  same  amount,  as  was  illustrated  in  §  26. 
Derived  from  these  are  the  following  useful  principles. 

(a)  A  term  may  be  transposed  (carried  over)  from  one  side 
(or  member)  of  an  equation  to  the  other  provided  its  sign  is 
changed. 

Thus,  in  the  equation  3  a:  —4  =2  we  may  transpose  the  term  —  4 
to  the  second  member,  giving  3  x  =2+4,  or  3  x  =6.  This  is  equiva- 
lent to  adding  4  to  both  members  of  the  given  equation. 

The  solving  of  equations  is  greatly  simplified  by  a  free  use 
of  this  principle  of  transposing  terms. 

Thus,  in  solving  3  x—  4=x+2,  we  may  transpose  the  —4  from 
the  first  member  to  the  second  and  at  the  same  time  transpose  the 
term  x  from  the  second  member  to  the  first,  giving  —  z+3  x  =2+4, 
or  2  x  =6.  Therefore  x  =3.  Ans. 

(b)  A  term  which  appears  in  both  members  of  an  equation 
may  be  canceled. 

Thus,  by  canceling  the  3  from  both  members  of  the  equation 
2  x  +3  =  10  +3,  we  have  simply  2  x  =  10,  and  hence  x  =5. 

Note  that  to  cancel  a  term  in  this  way  merely  amounts  to 
subtracting  it  from  both  members  of  the  given  equation. 

(c)  The  signs  of  all  the  terms  in  an  equation  may  be  changed. 
Thus  -5z+3=z-9may  be  written  5  x -3=  -x+9. 

Note  that  to  change  all  signs  in  this  manner  amounts 
to  multiplying  both  members  of  the  given  equation  by  —1. 

(d)  An  equation  may  be  cleared  of  fractions  by  multiplying 
both  members  by  the  lowest  common  denominator  of  all  the 
fractions. 


38  SECOND    COURSE    IN   ALGEBRA  [V,  §27 


EXAMPLE.     Solve  the  equation          -          =  6 


SOLUTION.     The  L.  C.  M.  of  the  denominators  is  12. 
Multiplying  both  members  by  12, 

4(z-2)  -3(3-3)  =72-6(s-l). 
Removing  parentheses  (§7), 

4z-8-3z+9=72-6z+6. 
Transposing,  6x+4:c_3  1=72+6+8_9> 

7*  =77. 
Therefore 

a:  =  11.     Ans. 

EXERCISES 

Solve  the  following,  using  the  principles  stated  in  §  27. 
Check  your  answer  for  the  first  five. 


2. 

!+6-T- 

33-1     2 

3-3  ,  6z+5 

/v.        0           C 
K         *         °           °    _Q 

2         3 

4            6 

2z      12 

1       3+1         2 

4          5 
7. 

8. 

Q 

5      8 
1             2 

6'   x       x*       33 

1 

3+1       3(3+1) 
1                    1 

12 

4-6y    2-3i 
1        ,r+l 

r-2 

r2_r_2     r-2     r+1 

^  rv  *k      I       -L  *v  "         O 

io-  jn-ija 


12 

' 


a;-4    x-8    z2-12z+32 


V,  §  27]  SIMPLE    EQUATIONS  39 

13.  If  10  be  subtracted   from  a   certain  number,  three 
fourths  of  the  remainder  is  9.     What  is  the  number? 

14.  Divide  38  into  two  parts  whose  quotient  is  122-. 

15.  Divide  96  into  two  parts  such  that  f  of  the  greater 
shall  exceed  f  of  the  smaller  by  6. 

16.  A  man  started  on  a  journey  with  a  certain  sum  of 
money.     He  spent  -J  of  it  for  car  fares  and  ^  of  it  for  hotel 
bills.     When  he  returned  home  he  found  he  had  $9.     How 
much  did  he  start  with  ? 

17.  I  have  $100  in  one  bank  and  $75  in  another  one.     If  I 
have  $45  more  to  deposit,  how  shall  I  divide  it  among  the  two 
banks  in  order  that  they  may  have  equal  amounts? 

18.  A  motor  boat  traveling  at  the  rate  of  12  miles  an  hour 
crossed  a  lake  in  10  minutes  less  tune  than  when  traveling 
at  the  rate  of  10  miles  an  hour.     What  is  the  width  of 
the  lake? 

[HiNT.     Time  =  Distance  -5-  Rate.] 

19.  A  freight  train  goes  6  miles  an  hour  less  than  a  passen- 
ger train.     If  it  goes  80  miles  in  the  same  time  that  a  passen- 
ger train  goes  112  miles,  find  the  rate  of  each. 

20.  A  tank  can  be  filled  by  one  pipe  in  10  hours,  or  by  an- 
other pipe  in  15  hours.     How  long  will  it  take  to  fill  the  tank 
if  both  pipes  are  open  ? 

[HINT.  Let  x  =  the  number  of  hours.  Then  l/x  =  the  part  both 
can  fill  in  1  hour.  But,  ^  =  the  part  the  first  pipe  can  fill  in  1  hour, 
and  T1g-=the  part  the  second  pipe  can  fill  in  1  hour.  Hence  we  must 

have  !  =  _1_L_JL~| 

x     10     15*  J 

21.  How  long  will  it  take  two  pipes  to  fill  a  tank  if  one 
alone  can  fill  it  in  5  hours  and  the  other  alone  in  12  hours? 


40  SECOND    COURSE    IN   ALGEBRA  [V,  §  27 

22.  Two  pipes  are  connected  with  a  tank.     The  large  one 
can  fill  it  in  3  hours ;   the  small  one  can  empty  it  in  4  hours. 
With  both  pipes  open,  how  long  before  the  tank  will  fill  ? 

23.  A  does  a  piece  of  work  in  4  days,  B  in  6  days,  and  C  in 
8  days.     How  long  will  it  take  them  working  together? 

24.  A  can  do  a  piece  of  work  in  16  hours,  and  B  can  do  it 
in  20  hours.     If  A  works  for  10  hours,  how  many  hours  must 
B  work  to  finish? 

25.  A's  age  is  ^  that  of  his  father's.     12  years  ago  he  was 
i  as  old  as  his  father.     How  old  is  each  now? 

26.  A  boat  goes  at  the  rate  of  12  miles  an  hour  in  still 
water.     If  it  takes  as  long  to  go  27  miles  upstream  as  45 
miles  downstream,  what  is  the  rate  of  the  current? 

27.  An  aviator  made  a  trip  of  95  miles.     After  flying  40 
miles,  he  increased  his  speed  by  15  miles  an  hour  and  made  the 
remaining  distance  in  the  same  time  it  took  him  to  fly  the 
first  40  miles.     What  was  his  rate  over  the  first  40  miles? 

28.  A  5-gallon  mixture  of  alcohol  and  water  contains  80% 
alcohol.     How  much  water  must  be  added  to  make  it  con- 
tain only  50%  alcohol? 

[HINT.     .50(z+5)=5X.80.     Explain.] 

29.  How  much  water  must  be  added  to  65  pounds  of  a 
10%  salt  solution  to  reduce  it  to  an  8%  solution  ? 

30.  A  train  660  feet  long  running  at  15  miles  an  hour  will 
pass  completely  through  the  Simplon  tunnel  in  Switzerland 
in  49^  minutes.    How  long  is  the  tunnel  ? 


DESCARTES 
(Rent  Descartes,  1596-1650) 

Profound  student  and  ranked  as  one  of  the  greatest  leaders  of  all  time  in 
both  mathematics  and  philosophy.  He  invented  representation  by  graphs 
and  was  thus  led  to  the  discovery  and  development  of  the  branch  of  mathe- 
matics called  Analytic  Geometry.  He  was  also  much  interested  in  medicine 
and  surgery. 


CHAPTER  VI 


GRAPHICAL   STUDY   OF  EQUATIONS 

28.  Definitions.  Let  two  lines  XX'  and  Y  Y'  be  drawn 
on  a  sheet  of  squared  (coordinate)  paper,  XX'  being  hori- 
zontal and  Y  Y'  vertical.  Two  such  lines  form  a  pair  of 
coordinate  axes.  The  point  0  where  they  intersect  is  called 
the  origin. 

Consider  any  point,  as  P,  and  draw  the  perpendiculars  PA 
and  PB  extending  from  P  to  the  two  axes  Y  Y'  and  XX' 
respectively.  PA  is  then  called  the  abscissa  of  P  and  PB 
is  called  the  ordinate  of  P.  The  abscissa  and  ordinate 
taken  together  are  called  the  coordinates  of  P. 


Y 

A 

A 

P 

i 

I 

(3 

4) 

(-2 

,3) 

n 

1 

X' 

- 

3 

0 

B 

X 

—  J 

4 

- 

2- 

1 

3  < 

) 

n 

S 

•*• 

(3 

) 

R 

-4 

( 

-3 

-4) 

Y' 

FIG.  5. 

Thus  the  point  P  in  the  figure  has  its  abscissa  equal  to  3  and  its 
ordinate  equal  to  4. 

All  abscissas  on  the  right  of   Y  Y'  are  considered  positive, 
while  all  abscissas  on  the  left  of  Y  Y'  are  considered  negative. 

Thus  the  abscissa  of  Q  is  -2 ;  that  of  R  is  -3  ;  that  of  S  is  +3. 

Similarly,  all  ordinates  above  XX'  are  considered  positive, 
while  all  ordinates  below  XX'  are  considered  negative. 

Thus  the  ordinate  of  Q  is  +3 ;  that  of  R  is  -4 ;  that  of  S  is  -2. 

41 


42  SECOND   COURSE    IN   ALGEBRA  [VI,  §  28 

In  reading  the  coordinates  of  a  point,  the  abscissa  is  always 
read  first  and  the  ordinate  second.  Thus,  in  the  figure,  the 
point  P  is  briefly  referred  to  as  the  point  (3,  4) ;  similarly,  Q 
is  the  point  (  — 2,  3)  ;  R  is  the  point  (  —  3,  —  4) ;  and  S  is  the 
point  (3,  —2),  etc. 

In  practice,  XX'  is  commonly  called  the  x-axis,  and  Y  Y' 
is  called  the  y-axis. 

EXERCISES  ' 

[The  pupil  will  find  it  convenient  to  use  the  prepared  coordinate 
paper  such  as  usually  may  be  secured  at  the  stationery  stores.] 

1.  Draw  axes  on  a  sheet  of  coordinate  paper  and  then 
locate  (plot)  the  following  points  : 

(2,4);   (-3,  -1);   (2,  -4);   (2J,  -3);   (-2J,  -2J); 
(-4,  +4);    (0,  -5);    (4,0);    (0,0). 

2.  The  part  of  the  plane  within  the  angle  XO  Y  (see  figure 
in  §  28)  is  called  the  first  quadrant,  the  part  within  the  angle 
YOX'  is  called  the  second  quadrant,  the  part  within  X'O  Y' 
is  called  the  third  quadrant,  etc.    Hence,  state  in  which  quad- 
rant a  point  lies  when 

(a)  its  abscissa  is  positive  and  its  ordinate  negative, 
(&)  its  abscissa  and  ordinate  are  both  negative, 
(c)  its  abscissa  is  negative  and  ordinate  positive. 

3.  What  can  be  said  of  the  position  of  a  point  whose  ordi- 
nate is  positive ;  whose  abscissa  is  negative  ? 

4.  A  certain  street  runs  due  east  and  west.     It  is  met  by 
another  street  which  runs  due  north  and  south,  thus  form- 
ing a  "  four  corners."     Taking  the  meeting  place  of  the  cen- 
ter-lines of  the  two  streets  as  origin,  and  the  east  and  north 
directions  as  positive,  what  are  the  coordinates  of  a  flagpole 
which  stands  due  northwest  from  the  origin  at  a  distance 
of  50  feet  from  the  center-line  of  each  road?    Answer  the 
same  when  the  pole  is  45  feet  due  west  of  the  crossing  point. 


VI,  §  29]     GRAPHICAL   STUDY   OP   EQUATIONS  43 

5.  Plot  the  following  three  points  and  then  see  if  it  is 
possible  to  draw  a  straight  line  that  will  pass  through  all  of 

them:  (1,5);  (0,3);  (-1,1). 

Do  the  same  for  the  three  points  (2,  3) ;  (—!,—!);  (5,  0). 

29.  Graph  of  an  Equation.  We  have  seen  in  Chapter  V 
that  if  we  have  any  linear  equation  containing  a  single  un- 
known letter,  as  for  example  the  equation  2  x—  l  =  3(x—  1), 
we  can  always  solve  it ;  that  is  we  can  find  the  value  of  x. 

Suppose  now  that  we  have  a  linear  equation  in  which 
two  unknown  letters,  x  and  y,  appear,  that  is  an  equation 
in  which  no  term  contains  both  x  and  y  nor  any  higher  power 
of  either  of  them  than  the  first,  as  for  example 
(1)  x+y  =  5. 

The  meaning  of  such  an  equation  and  the  interesting 
facts  about  it  are  best  brought  out  by  graphical  methods  in 
ways  which  we  shall  now  explain. 

In  the  first  place,  it  is  to  be  observed  that  such  an  equa- 
tion is  satisfied  by  a  great  many  pairs  of  values  for  x  and  y. 
For  example,  the  pair  of  values  0=1,  y  =  4)  satisfies  the 
equation,  because  when  we  put  these  values  for  x  and  y 
respectively  in  the  equation,  it  becomes  1+4  =  5,  which  is 
true.  Again,  the  same  is  seen  to  be  true  of  the  pair  (x  =  2, 
2/  =  3)  (explain) ;  and,  similarly,  the  same  is  true  of  any  one  of 
the  pairs  (x  =  ±,  y  =  f),  (x  =  Q,  y=-l),  (x  =  8,  y=-3),  etc. 
In  fact,  we  can  obtain  as  many  such  x,  y  pairs  as  we  wish, 
each  pair  having  the  property  that  the  z-value  and  the 
?/-value  taken  together  satisfy  the  given  equation. 

If  we  place  x  =3  in  the  equation  above,  we  have  3  +y  =5  and  this, 
when  solved  for  y,  gives  y  =2.  Thus  (x  =3,  y  =2)  is  a  pair  such  as 
mentioned  above.  Similarly,  we  can  assign  to  x  any  value  we  wish 
(positive  or  negative)  and  find  from  the  equation  the  corresponding 
value  of  y,  thus  forming  a  new  pair  of  values  of  x  and  y. 


44 


SECOND   COURSE    IN  ALGEBRA          [VI,  §29 


Whenever  an  equation  contains  two  (but  no  more)  un- 
known letters,  such  as  x  and  y,  any  pair  of  values  for  x  and 
y  that  satisfy  it  is  called  a  solution  of  the  given  equation. 
It  follows  from  what  has  been  shown  above  that  every  such 
equation  has  an  indefinitely  large  number  of  solutions. 

Returning  to  the  equation  x-\-y  =  5,  let  us  consider  again 
the  special  solutions  which  we  noticed  on  page  43 : 
(x  =  2, 2/  =  3),  (z  =  i,  2/  =  f),  (x=-l,  i/  =  6),  (x  =  8,  2/=-3). 
Following  the  ideas  brought  out  in  §  28,  each  of  these  may 
now  be  plotted  a*s  a  point,  using  x  as  abscissa  and  y  as  ordi- 
nate.  Upon  locating  these  points  carefully,  it  will  be  seen 
that  they  all  lie  on  one  and  the  same  straight  line,  as  indi- 
cated in  the  figure  below. 


\ 

Y 

\ 

\< 

i- 

-) 

\ 

(2 

3) 

•+ 

X 

A 

\ 

0 

\ 

X 

\ 

(6 

-U 

\ 

\ 

( 

8, 

3) 

\ 

FIG.  6. 

This  line  is  called  the  graph  of  the  equation  x+y  =  5.  It 
may  be  shown  that  every  solution  of  the  given  equation  gives 
rise  when  plotted  to  some  point  upon  this  line,  and  vice  versa, 
every  point  upon  this  line  has  an  a>value  and  a  i/-value 
which,  when  taken  together,  form  a  solution  of  the  given 
equation. 

30.  Graph  Determined  from  Two  Points.  In  practice, 
the  graph  of  a  linear  equation  is  drawn  by  locating  two  points 
upon  it,  and  connecting  them  by  a  straight  line. 


VI,  §  30]     GRAPHICAL   STUDY   OF   EQUATIONS 


45 


EXAMPLE.     Draw  the  graph  of  the  equation  5  x— 4  y  =  20. 

SOLUTION.  Placing  x  =0  in  the  equation  gives  y  =  —  5.  Hence 
(0,  —5)  is  a  point  on  the  graph. 

Placing  y  =  0  in  the  equation  gives  x  =  4.  Hence  (4,  0)  is  a  point 
on  the  graph. 

Plotting  these  two  points  (0,  —5)  and  (4,  0)  and  drawing  (with 
ruler)  the  straight  line  through  them,  gives  the  required  graph. 


Y 

V 

A 

f 

/ 

0 

(4 

0) 

/ 

x 

/ 

/ 

^ 

/ 

« 

4 

/ 

/ 

(0 

-5 

) 

b/ 

FIG.  7. 

NOTE.  As  in  the  example  just  considered,  it  is  often  simplest 
to  select  as  the  first  point  the  one  whose  abscissa  is  0,  and  as  the 
second  point  the  one  whose  ordinate  is'O.  However,  it  is  equally 
correct  to  take  any  two  points  whose  coordinates  satisfy  the  given 
equation.  If  the  two  points  selected  are  too  close  to  each  other, 
it  is  difficult  to  draw  the  line  accurately ;  if  this  happens,  plot  a  third 
point  on  the  line  at  a  considerable  distance  from  the  first  two. 

EXERCISES 

1.  State  (orally)  what  is  true  of  each  point  on  the  line  in 
the  last  figure. 

[HINT.  Its  abscissa  and  ordinate,  taken  as  a  pair  of  numbers, 
(x,  y),  form  a  .  .  .  .] 

Draw  the  graph  of  each  of  the  following  linear  equations. 

2.  2x-y  =  4:.  4.   2x+3y=12.  6.   4x  =  3y-7. 

3.  2x+y  =  2.  5.   x-3y  =  3.  1.   2x  =  3y. 


46  SECOND   COURSE   IN   ALGEBRA  [VI,  §  30 

8.  If  a  person  travels  at  the  rate  of  15  miles  per  hour,  the 
distance  s  which  he  will  have  traveled  at  the  end  of  t  hours 
is  given  by  the  formula  s  =  15  t.     Draw  the  graph  of  this 
equation,  using  the  ^-values  as  abscissas  and  the  s-values  as 
ordinates.     From  your  figure  read  off  (approximately)  how 
far  he  will  have  traveled  at  the  end  of   (a)  2  hours ;    (6)  3 
hours;  (c)  3^-  hours;  (d)  4-J-  hours. 

[HINT.  Take  the  Z-axis  horizontal  and  the  s-axis  vertical,  and 
let  the  unit  length  on  each  be  about  half  an  inch.  In  order  to  get 
the  diagram  into  relatively  compact  form,  allow  each  unit  on  the 
s-axis  to  represent  15  miles,  taking  each  unit  on  the  2-axis  to  repre- 
sent 1  hour.] 

9.  A  boy  has  $10  in  the  bank  when  he  begins  saving  at  the 
rate  of  $3  a  month,  adding  this  amount  month  by  month  to 
his  account.      Find  graphically  how  many  months  must 
elapse  before  his  account  will  amount  to  $22. 

[HINT.  Let  A  represent  the  amount  of  the  account  at  the  end  of 
t  months.  Then,  A  =10  +3  L  (Why?)  Now  draw  the  graph  of 
this  equation,  using  ^-values  as  abscissas  and  A -values  as  ordinates, 
and  taking  for  convenience*  one  unit  on  the  A -axis  to  represent  $2, 
while  one  unit  on  the  t-axis  represents  1  month.  The  problem  then 
calls  for  that  abscissa  which  goes  with  the  ordinate  A  =22.] 

10.  A  boy  has  $30  in  the  bank  when  be  begins  spending  it 
at  the  rate  of  $4  a  month.     Find  graphically  how  long  it 
will  be  before  he  has  but  $2  left. 

[HINT.     Use  the  same  letters  and  units  as  in  Ex.  9.] 

11.  A  wheel  is  rotating  at  the  rate  of  10  revolutions  a 
second  when  the  power  is  shut  off.     The  wheel  slows  down 
uniformly  and  comes  to  rest  at  the  end  of  30  seconds.     Make 
a  diagram  from  which  you  can  read  off  how  many  revolutions 
the  wheel  was  making  at  any  given  instant  after  the  power 
was  shut  off  and  use  your  diagram  to  determine  how  many 


VI,  §  31]     GRAPHICAL   STUDY    OF   EQUATIONS 


47 


revolutions  per  second  were  being  made  at  the  end  of  (a)  6 
seconds ;   (6)  9  seconds ;   (c)  18  seconds ;  (d)  26  seconds. 

[HINT.  Let  r  represent  the  number  of  revolutions  per  second  at 
the  end  of  t  seconds.  Then  the  conditions  of  the  problem  tell  us 
that  r  =  10  when  t=0,  and  that  r=0  when  £=30.  Thus  we  have 
two  points  on  the  graph,  and  we  can  draw  the  graph  completely 
without  even  getting  its  equation.] 

12.  The  temperature  at  which  water  freezes  is  32°  on  the 
Fahrenheit  scale,  but  it  is  0°  on  the  Centigrade  scale.  The 
temperature  at  which  water  boils  is  212°  on  the  Fahrenheit 
scale,  but  it  is  100°  on  the  Centigrade  scale.  Make  a  dia- 
gram from  which  you  can  read  off  the  Centigrade  tempera- 
ture that  corresponds  to  any  given  Fahrenheit  temperature. 

[HINT.  Let  F  represent  the  Fahrenheit  reading  and  C  the  Centi- 
grade reading.  Then  C=0  when  F=  32  and  C=100  when  F=212. 
This  gives  two  points.  The  graph  is  the  straight  line  joining  these 
two  points.] 

31.  Simultaneous  Equations.  Suppose  that,  instead  of 
having  a  single  linear  equation  containing  the  two  unknown 
letters  x  and  y  (as  in  §  29),  we  have  two  such  equations; 
for  example 

x+y  =  Q  and  3  x-2  y=  -2. 
Of  all  the  pairs  of  values  (x,  y)  that 
will  satisfy  the  first  equation  and  all 
the  pairs  (x,  y)  that  will  satisfy  the 
second  equation,  is  there  a  particu- 
lar pair  (x,  y)  that  will  satisfy  them 
both  at  the  same  time?  We  shall 
consider  this  question  graphically. 

Draw  the  graphs  of  the  two  equa- 
tions on  the  same  sheet  of  coordinate  paper,  using  the  same 
axes  throughout.  The  lines  thus  obtained  are  seen  to  in- 
tersect each  other  in  the  point  (2,  4).  This  means  that  the 


\ 


FIG.  8. 


48 


SECOND   COURSE   IN  ALGEBRA          [VI,  §  31 


pair  (x  =  2,  2/  =  4)  satisfies  both  equations  at  once,  since  it 
lies  on  both  the  graphs.  This  pair  (x  =  2,  y  =  4)  is  there- 
fore the  pair  desired,  and  it  is  the  only  such  pair  because 
two  straight  lines  can  intersect  in  but  one  point. 

That  this  answer  is  correct  is  readily  seen  by  substituting  this 
pair  of  values  in  the  given  equations.  Thus,  with  x=2  and  y  =  4, 
the  equations  become  2+4  =  6  and  6  —  8=  —2,  which  are  true. 

The  two  equations  above  illustrate  what  is  known  as  a 
set  of  simultaneous  equations,  and  the  particular  pair  of 
values  (x  =  2,  i/  =  4)  which  we  found  would  satisfy  both  the 
equations  at  one  time,  illustrates  what  is  called  the  solution 
of  the  set.  In  general,  two  or  more  equations  are  said  to  be 
simultaneous  if  they  are  considered  at  the  same  time.  In 
the  present  chapter  we  shall  deal  only  with  sets  containing 
two  unknowns,  as  in  the  preceding  example. 

32.  Inconsistent  Equations.  Although  two  linear  simul- 
taneous equations  in  x  and  y  will  in  general  have  a  solution 
(as  in  §  31),  there  are  cases  where  no 
solution  can  be  found,  and  indeed 
none  exists.  For  example,  if.  we  draw 
the  graphs  of  the  equations 

x+y  =  3,  and  x+y  =  6, 
we  see  that  the  lines  do  not  intersect ; 
in  other  words,  they  are  parallel. 
Thus,  there  is  no  pair  of  values  (x,  y) 
that  will  satisfy  both  equations  at 
once ;  that  is,  there  is  no  solution.  Such  a  pair  of  simul- 
taneous equations  is  called  inconsistent. 

EXERCISES 

Determine  graphically  which  of  the  following  sets  of  simul- 
taneous equations  has  a  solution  arid  which  does  not.  In 


\ 

Y 

\ 

\ 

\ 

a 

\ 

Xk 

X 

t 

\ 

\ 

o3 

\ 

\ 

\ 

X 

0 

\ 

\ 

\ 

"^ 

u>  •>  *  * 

VI,  §32]     GRAPHICAL   STUDY   OF   EQUATIONS         ,      49 

case  a  solution  exists,  determine  it  and  check  your  result  by 
substituting  it  in  the  given  equations. 

f 


9.  A  man  starts  at  a  given  time  and  walks  along  a  certain 
road  at  the  rate  of  5  miles  an  hour.  An  hour  later  another 
man  starts  from  the  same  place  and  travels  in  the  same 
direction  at  the  rate  of  10  miles  an  hour.  Find  (graphically) 
how  far  from  the  starting  point  they  will  meet. 

[HINT.  If  s  represents  distance  (in  miles)  traveled  in  t  hours, 
the  first  man's  motion  is  described  by  the  equation  s  =5  t,  while  the 
second  man's  motion  is  described  by  the  equation  s  =  10(Z  —  1). 
Now  draw  a  pair  of  axes,  and  draw  in  the  graphs  of  these  two  equa- 
tions, using  lvalues  as  abscissas.  The  problem  then  calls  for  that 
value  of  s  which  belongs  to  the  intersection  of  the  two  graphs.] 

10.  Use  your  diagram  for  Ex.  9  to  answer  the  following 
question  :  After  how  much  time  will  the  two  men  meet  ? 

11.  A  man  starts  and  walks  along  a  certain  road  at  the 
rate  of  5  miles  an  hour.     At  the  same  instant  another  man 
starts  out  at  a  point  on  the  same  road  15  miles  distant  and 
travels  toward  the  first  man  on  a  bicycle  at  the  rate  of  10 
miles  an  hour.     How  far  from  the  first  man's  starting  point 
will  they  meet?     How  long  will  it  take  them? 

12.  B  and  C  start  to  save  money.     B  has  $10  when  they 
begin  and  saves  at  the  rate  of  $3  a  month,  while  C  at  the 
start  owes  $6  and  saves  at  the  rate  of  $7  a  month.     Find 
graphically  how  soon  C  will  be  able  to  cancel  his  debt  and 
have  savings  equal  to  B's,  and  how  much  each  will  then  have. 


CHAPTER  VII 

SIMULTANEOUS    EQUATIONS    SOLVED    BY 
ELIMINATION 

33.  Elimination  by  Substitution.  The  process  of  combin- 
ing two  equations  in  two  unknowns  in  such  a  way  as  to 
cause  one  of  the  unknowns  to  disappear  is  called  elimina- 
tion. 

We  shall  consider  first  the  method  called  elimination  by 
substitution.  The  process  is  illustrated  by  the  following 
example. 

EXAMPLE.     Solve  the  system 

(1)  2x+3y  =  2, 

(2)  5z-4i/  =  28. 

SOLUTION.     From  (1), 

(3)  2x=2-3y. 
Therefore 


(4) 


o  o 

Substituting  ^  for  x  in  (2)  gives 


(5) 


Clearing  (5)  of  fractions, 

(6)  5(2  -3  y)  -8  y  =  56. 

50 


VII,  §  34]  ELIMINATION                                      51 

Simplifying, 

(7)  W-15y-8y=5G. 
Collecting, 

(8)  -23  y  =  46. 
Therefore 

(9)  y=-2. 
Substituting  -2  for  y  in  (4)  now  gives 


The  required  solution  of  the  system  (1),  (2)  is,  therefore, 

(z=4,  i/=-2).     Ans. 

CHECK.     Substituting  x  =4  and  y  =  -2  in  (1)  gives 

2(4)+3(-2), 

which  is  equal  to  8  — 6,  or  2,  as  (1)  requires. 
Substituting  x  =4  and  y  =  —  2  in  (2)  gives 

5(4) -4( -2), 
which  is  equal  to  20+8,  or  28,  as  (2)  requires. 

To  solve  two  simultaneous  equations  by  substitution  : 

1.  Solve  either  equation  for  one  of  the  unknown  letters  in 
terms  of  the  other  one. 

2.  Place  the  result  thus  obtained  in  the  other  equation  and 
solve  it. 

3.  Having  thus  found  one  of  the  unknown  letters,  substitute 
its  value  in  either  of  the  given  equations  and  solve  for  the  other 
unknown  letter. 

34.  Elimination  by  Addition  or  Subtraction.  The  only 
other  method  of  elimination  which  we  shall  consider  here  is 
called  elimination  by  addition,  or  subtraction.  The  process  is 
illustrated  by  the  following  example. 


52  SECOND   COURSE   IN  ALGEBRA        [VII,  §  34 

EXAMPLE.     Solve  the  system 

(1)  3s+4y=12, 

(2)  2z-5y  =  54. 

SOLUTION.     Multiplying  (1)  by  2, 

(3)  6z+8'2/=24. 
Multiplying  (2)  by  3, 

(4)  6z-15  y  =  162. 
Subtracting  (4)  from  (3) , 

(5)  232/=-138. 
Therefore 

2/=-6. 

Substituting  y  =  -6  in  (1),  3  x  -24  =  12,  or,  3  x  =36. 

Therefore  x  =  l2. 

The  required  solution  of  the  system  (1),   (2)  is,  therefore, 

(z  =  12,  y=  —6).     Ans. 
CHECK.     Substituting  1?  for  x  and  —6  for  y  in  (1),  gives 

3(12) +4( -6)  =36 -24= 12,  as  (1)  requires. 
Substituting  12  for  x  and  —6  for  y  in  (2),  gives 

2(12)  -5(  -6)  =24+30=54,  as  (2)  requires. 

NOTE.  Instead  of  multiplying  (1)  by  2  and  (2)  by  3  and  then 
subtracting  them,  thus  eliminating  x,  we  might  just  as  well  have 
multiplied  (1)  by  5  and  (2)  by  4  and  added  them,  thus  eliminating  y. 
Either  plan  leads  to  the  same  solution  for  the  given  system. 

To  solve  two  simultaneous  equations  by  addition  or  subtrac- 
tion: 

1.  Multiply  one,  or  both,  of  the  given  equations  by  such  num- 
bers as  will  make  the  coefficients  of  one  of  the  letters  (say,  y) 
numerically  equal. 

2.  Subtract  (or  add)  the  two  equations  thus  obtained,  thus 
eliminating  one  of  the  unknown  letters. 

3.  Solve  the  resulting  equation  for  the  letter  it  contains,  and 
obtain  the  value  of  the  other  letter  by  substituting  the  value  of 
the  letter  already  found  into  either  of  the  given  equations. 


VII,  §  34] 


ELIMINATION 
EXERCISES 


53 


1. 


Solve  by  substitution : 
2x+Zy=12, 
x+5y=13. 


3. 

4.    < 


[2x-3y=-W. 

5.    <           _ ,  / 

\                ^*         ^  4  * 

Solve  by  addition  or  subtraction : 


6. 


[        A+B=-9,                     {10*-3V—  .6, 

\7A-3B  =  7.                          \    7x+42/  =  8. 

Solve  by  either  method  : 

x    4  y 

3  x  —  y    x-\-y 

11. 

9         Q            J 

2                                  15' 

2~        3          ' 
3z+i/    3o:-2/_     2 

3 

11            3 

[HINT.     First  clear  of  fractions.] 

12. 

2H-6,  s_8 

»i.  r  ' 

M-n, 

x     y 
10     12     4 

12        L 

—  4:. 

HINT  TO  Ex.  16.   Do  not  clear  of  fractions,  but  solve  for  -  and  --~| 

2x-5?/=-l, 

-+—  =  5, 

13. 

x-2/    3x+2y 
7            23 

15_30=_1 
«      V 

14. 

6x-8i/    8x-20i/    ft 

1     1      1       5 

0                                  11                           ' 

.                                18' 

x+V=|- 

i    r      .   ^       £>> 

x-1     2/+1 
2     ,     3    _10 

X—l        y+l 

54  SECOND    COURSE   IN   ALGEBRA        [VII,  §  35 

35.  Simultaneous  Equations  in  Three  Unknown  Letters. 
We  often  meet  with  a  system  of  three  linear  equations  between 
three  unknown  letters.  Such  a  system,  like  those  already 
considered  (§§  33,  34),  may  be  solved  by  elimination. 

EXAMPLE.     Solve  the  system 

(1)  x+y+z  =  6, 

(2)  2z-2/+3z  =  9, 

(3)  x+2y-z  =  2. 

SOLUTION.  Eliminate  one  of  the  unknowns,  say  y,  from  (1) 
and  (2).  Thus 

(4)  3  x  +42  =  15.     [(l)+(2)] 
Eliminate  the  same  unknown,  y,  from  (2)  and  (3).     Thus 

(5)  4  x  -2  y  +6  z  =  18.     (2)X2 

(6)  x+2y-    2  =  2.     (3) 

5x          +52=20,     (5) +  (6) 
or 

(7)  x  +2  =  4. 

Equations  (4)  and  (7)  contain  only  x  and  z  and  hence  may  be 
solved  for  these  letters,  as  in  §§  33,  34.  Thus 

(8)  3  x  +4  z  =  15.     (4) 

(9)  3  x  +3  z  =  12.     (7)  X3 

0=3.       (8) -(9) 

Substituting  z  =3  in  (7),  we  obtain  z+3  =4.     Therefore  x  =  1. 
Substituting  2=3  and  z  =  1  in  (1),  we  find  l+y+3  =6.     There- 
fore y=2. 

The  desired  solution  is,  therefore,  (x  =  l,  y  =2,  2=3).     Ans. 

To  solve  three  simultaneous  equations : 

1.  Eliminate  one  of  the  unknown  letters  from  any  pair  of 
the  equations,  then  eliminate  the  same  unknown  from  any  other 
pair  of  the  equations. 

2.  Solve  the  two  equations  thus  obtained,  as  in  §  34. 

3.  This  gives  two  of  the  letters,  and  the  third  may  then  be 
found  by  substituting  the  letters  already  found  in  either  of  the 
given  equations. 


-  >  a 


VII,  §35] 


H 


ELIMINATION 


\\VA--  tA\-u 
^vr-  to 


55 


EXERCISES 

Solve  for  x,  y,  and  z  each  of  the  following  sets  of  equations. 

+3^  =  14,  \x+y-z  =  Q, 

1.          2z+?/+22  =  10,  4.  z-?/  =  4, 

-3^  =  2.  z+z  =  7. 


2. 


3. 


5. 


M+i-2, 

x    y     z 

2-i+i  =  7, 

x     ?/     z 


[HINT.     See  hint  to  Exer- 
cise 16,  p.  53.] 


APPLIED    PROBLEMS 

1.  The  sum  of  two  numbers  is  75  and  their  difference  is  5. 
Find  the  numbers. 

[HINT.     Let  x  be  one  of  the  numbers  and  y  the  other,  and  form 
two  equations.] 

2.  One  third  of  the  sum  of  two  numbers  is  10,  while  one 
sixth  of  their  difference  is  1.     Find  the  numbers. 

3.  The  perimeter  of  a  certain  rectangle  is  10  inches  less 
than  3  times  the  base.     If  the  base  is  4|-  times  the  height, 
what  are  the  base  and  height? 

4.  Each  base  angle  of  a  certain  isosceles  triangle  is  66° 
more  than  the  vertical  angle.     Find  each  angle. 

5.  A  father's  age  is  1^-  that  of  his  son.     Twenty  years  ago 
his  age  was  twice  his  son's.     How  old  is  each? 

6.  Four  years  ago  A's  age  was  2^  B's  age.     Four  years 
hence  A's  age  will  be  1T8T  B's  age.     What  is  the  age  of 
each? 


56  SECOND   COURSE   IN  ALGEBRA        [VII,  §  35 

7.  A  part  of  $2500  is  invested  at  6%  and  the  remainder 
at  5%.     The  yearly  income  from  both  is  $141.     Find  the 
amount  in  each  investment. 

8.  One  sum  of  money  is  invested  at  5%  and  another  at 
6%.     The  total  yearly  income  from  both  investments  is 
$53.75.     If  the  rates  should  be  reversed,  the  annual  income 
would  be  increased  by  $2.50.     Find  the  sums  of  money  in- 
vested. 

9.  A  and  B  together  can  do  a  piece  of  work  in  12  days. 
After  A  has  worked  alone  for  5  days,  B  finishes  the  work  in 
26  days.     In  what  time  could  each  do  the  work  alone? 

[HINT.  Let  x  =  the  time  in  which  A  can  do  it  alone,  y  =  the  time 
in  which  B  can  do  it  alone.  Then  the  part  A  can  do  in  one  day  is  -, 
and  the  part  B  can  do  in  one  day  is  -.  So  the  equations  become 

i  +1  =  J_  and  5  +?§  =  l.     Now  solve  as  in  Ex.  16,  p.  53.] 
x     y     12          x      y 

10.  A  and  B  can  do  a  certain  piece  of  work  in  16  days. 
They  work  together  for  4  days,  when  B  is  left  alone  and 
completes  the  work  in  36  days.     In  what  time  could  each 
do  it  separately? 

11.  A  laborer   agreed  to   stay  on  a   farm  for  100  days. 
For  each  day  he  worked  he  was  to  receive  $2  and  board, 
but  for  each  idle  day  he  was  to  forfeit  75  cents  for  his  board. 
When  the  time  expired,  he  received  $180.75.     How  many 
days  did  he  work? 

12.  An  errand  boy  went  to  the  bank  to  deposit  some  bills, 
some  of  them  being  $1  bills  and  the  rest  $2  bills.     If  there 
were  38  bills  in  all  and  their  combined  value  was  $50,  how 
many  of  each  were  there? 

[HINT.  Let  rr=the  number  of  $1  bills,  and  y  =the  number  of 
$2  bills.  Then  their  combined  value  was  x  +2  y  dollars.] 


VII,  §  35]  ELIMINATION  57 

13.  The  receipts  from  the  sale  of  300  tickets  for  a  musical 
recital  were  $125.     Adults  were  charged  50  cents  each,  and 
children  25  cents  each.     How  many  tickets  of  each  kind  were 
sold? 

14.  A  grocer  wishes  to  make  50  pounds  of  coffee  worth 
32  t;ents  per  pound  by  mixing  two  other  grades,  one  of  which 
is  worth  26  cents  per  pound  and  the  other  35  cents  per  pound. 
How  much  of  each  must  he  use  ? 

15.  One  cask  contains  18  gallons  of  vinegar  and  12  gallons 
of  water;     another,  4  gallons  of  vinegar  and  12  of  water. 
How  many  gallons  must  be  taken  from  each  so  that  when 
mixed  there  may  be  21  gallons,  half  vinegar  and  half  water? 

16.  Two  cities  are  140  miles  apart.     To  travel  the  distance 
between  them  by  automobile  takes  3  hours  less  time  than  by 
bicycle,  but  if  the  bicycle  has  a  start  of  42  miles,  each  takes 
the  same  time.     What  is  the  rate  of  the  automobile,  and 
what  the  rate  of  the  bicycle  ? 

17.  A  boy  rows  18  miles  down  a  river  and  back  in  12  hours. 
He  can  row  3  miles  downstream  while  he  rows  but  1  mile 
upstream.     What  is  his  rate  in  still  water,  and  what  is  the 
rate  of  the  stream  ? 

18.  A  motor  boat  can  run  r  miles  an  hour  in  still  water. 
If  it  went  downstream  for  s  hours  and  took  t  hours  to  return, 
what  was  the  total  distance  traveled,  and  what  was  the  rate 
of  the  current? 

19.  The  sum  of  three  numbers  is  20.     The  sum  of  the  first 
and  second  is  10  greater  than  the  third,  while  the  difference 
between  the  second  and  third  is  6  less  than  the  first.     Find 
the  numbers. 

[HINT.     Use  the  three  letters  x,  y,  z  to  represent  the  unknown 
numbers,  and  form  three  equations.     Solve  as  in  §  35.] 


58  SECOND   COURSE    IN   ALGEBRA        [VII,  §  35 

20.  A,  B,  and  C  have  certain  sums  of  money.     B  would 
have  the  same  as  A  if  A  gave  him  $100 ;   C  would  have  four 
times  as  much  as  B  if  B  gave  him  $100 ;  and  C  would  have 
twice  as  much  as  A  if  A  gave  him  $100.     How  much  has 
each? 

21.  I  have  $90  on  deposit  in  bank  A,  $51  in  bank  B,  and 
$75  in  bank  C.     If  I  have  $144  more  to  deposit,  how  shall  I 
distribute  it  among  the  three  banks  so  as  to  make  the  three 
deposits  equal? 

22.  The  perimeter  of  a  certain  rectangle  is  16  feet.     If 
the  length  be  increased  by  3  feet  and  the  breadth  by  2  feet, 
the  area  becomes  increased  by  25  square  feet.     What  are 
the  length  and  breadth? 

23.  A  barrel  of  vinegar  is  to  be  bottled  for  selling  and  it  is 
desired  that  some  of  the  bottles  be  of  pint  size,  others  of 
quart  size  and  ojthers  of  gallon  size.     In  order  that  there  be 
52  bottles  in  all,  and  twice  as  many  of  the  pint  as  of  the  quart 
size,  how  many  of  each  will  be  necessary? 

[HINT.     1  barrel  =32  gallons.] 

24.  For  any  pulley  block,  the  relation  between  the  weight 
to  be  raised  and  the  pull  necessary  to  raise  it  is 

pull  =  x+yX  weight, 

where  x  and  y  are  numbers  that  are  different  for  different 
pulleys. 

In  two  experiments  with  a  certain  pulley  block,  a  weight 
of  100  pounds  was  raised  by  a  pull  of  22  pounds,  and  a  weight 
of  200  pounds  was  raised  by  a  pull  of  42  pounds.  Find  the 
values  of  x  and  y  for  this  pulley. 


CHAPTER  VIII 
SQUARE   ROOT 

36.  Definitions.     The  square  root  of  a  given  number  is 
the  number  whose  square  equals  that  number. 

Thus  2  is  the  square  root  of  4  because  22  =4.     Likewise,  3  is  the 
square  root  of  9  because  32  =9,  etc. 

The  square  root  of  a  number  is  denoted  by  the  radical  sign 
V      placed  over  it. 

Thus    V4=2,   V9=3,   Vl6=4,  etc. 

The  process  of  finding  the  square  root  of  a  number  is  called 
extracting  its  square  root. 

37.  Extracting  Square  Roots  in  Arithmetic.     Many  times 
we  can  pick  out  the  square  root  of  a  number  by  inspection. 
Thus,  A/Ill  is  seen  to  be  12  because  122  =  144.    Similarly, 
Vl96  =  14.     But  in  finding  the  square  root  of  a  large  number, 
such  as  74,529,  we  cannot  ordinarily  determine  the  answer 
by  mere  inspection.     The  process  for  such  a  case  is  illustrated 
below,  and  is  explained  on  the  next  page. 

PROCESS. 

7'45'29  |  273      Ans. 
4 
Trial  divisor         =2x20   =40 


Complete  divisor  =  40 +7   =47 


345 
329 


Trial  divisor         =  2  X270  =  540 
Complete  divisor  =  540+ 3  =  543 
59 


1629 
1629 


60  SECOND   COURSE    IN   ALGEBRA      [VIII,  §  37 

EXPLANATION.  First  separate  the  number  into  periods  of  two 
figures  each,  beginning  at  the  right.  That  is,  in  the  present  case, 
write  the  number  in  the  form  7 '45 '29. 

Find  the  greatest  square  in  the  left-hand  period  and  write  its 
root  for  the  first  figure  of  the  required  root.  This  gives  the  2  ap- 
pearing in  the  answer. 

Square  this  figure  (giving  4),  subtract  the  result  from  the  left- 
hand  period  and  annex  to  the  remainder  the  next  period  for  new 
dividend.  This  gives  the  345  appearing  in  the  process. 

Double  the  root  already  found,  with  a  0  annexed  (giving  40)  for 
a  trial  divisor  and  divide  the  last  dividend  (345)  by  it.  The  quo- 
tient (or,  in  some  cases,  the  quotient  diminished)  forms  the  second 
figure,  7,  of  the  required  root.  Add  to  the  trial  divisor  the  figure 
last  found  (7),  giving  the  complete  divisor  (47).  Multiply  this 
complete  divisor  by  the  figure  of  the  root  last  found  (7),  giving  the 
329  appearing  in  the  process.  Subtract  this  from  the  dividend,  and 
to  the  remainder  annex  the  next  period  for  the  next  dividend.  This 
gives  the  1629  of  the  process. 

Proceed  as  before,  and  continue  until  a  new  dividend  equal  to  0 
is  obtained.  In  the  example  above,  this  happens  at  once,  giving 
273  as  the  required  root. 

This  process  is  the  one  commonly  used  in  arithmetic,  and 
is  stated  here  as  a  review.  We  shall  see  in  §  38  that  a  sim- 
ilar process  may  be  used  in  extracting  the  square  roots  of 
expressions  in  algebra. 

In  the  example  just  solved,  the  root  comes  out  exact  be- 
cause 74,529  (whose  root  is  being  extracted)  is  a  perfect 
square  —  that  is,  it  is  like  one  of  the  numbers  1,  4,  9,  16,  36, 
etc.  If  we  had  started  with  a  number  which  was  not  a  per- 
fect square,  the  process  would  be  the  same  except  that  we 
should  not  finally  reach  a  new  dividend  which  equals  0. 
In  such  cases,  in  fact,  the  process  continues  indefinitely,  but 
if  we  stop  it  at  any  point,  we  have  before  us  the  desired  root 
correct  (decimally)  up  to  that  point.  For  example,  in  find- 
ing the  square  root  of  550  correct  to  two  decimal  places,  the 
process  is  as  follows. 


VIII,  §  37]  SQUARE   ROOT  61 

PROCESS. 


5'50.00'00  |  23.45      Ans. (correct  to  two 
4  decimal  places) 


2X20  =  40 
40+3  =  43 


150 
129 


2X230  =  460 
460+4  =  464 


2100 
1856 


2X2340  =  4680 
4680+5  =  4685 


24400 
23425 


975 

NOTE.  In  the  process  above  we  have  first  written  550  in  the  form 
550.0000.  If  we  had  written  it  with  six  zeros,  that  is  550.000000, 
and  then  carried  the  process  forward  until  all  these  were  used  be- 
low, we  should  have  obtained  the  root  correct  to  three  decimal 
places  instead  of  two.  In  general,  the  root  obtained  would  be 
correct  to  a  number  of  decimal  places  equal  to  half  the  number  of 
zeros  added. 

Square  roots  of  decimal  numbers,  such  as  334.796,  are  ob- 
tained like  those  for  whole  numbers,  except  that  in  the  begin- 
ning the  separation  of  the  number  into  periods  of  two  figures 
each  must  be  carried  out  both  ways  from  the  decimal  point. 

Thus  334.796  would  be  written  3 '34.79 '60.  Similarly,  3.67893 
would  be  written  3 '.67 '89 '30.  The  extraction  of  the  root  is  then 
carried  out  as  in  the  process  shown  above. 

EXERCISES 

Find  (by  inspection  or  by  the  process  shown  in  §  37)  the 
square  root  of  each  of  the  following  numbers. 

1.   49.  5.   576.  9.   8281.          13.   f 

=  |Xf.] 


2.  81. 

6.  1444. 

10.  15,876. 

[HINT. 

3.  64. 

7.  4225. 

11.  42,025. 

14-  M 

4.  169. 

8.  1681. 

12.  95,481. 

15.  & 

62                    SECOND    COURSE    IN  ALGEBRA       [VIII,  §  37 

Find  the  square  root  of  each  of  the  following  numbers 
correct  to  two  decimal  places. 

16.  567.                     19.    17.76.  22.   3. 

17.  633.                     20.    13.  23.   f. 

[HINT.  Write  as  [HINT.  Write    f    as 

13'.00'00.]  .75.] 

18.  1305.                  21.   2.  24.        . 


38.   Extracting  Square  Roots  in  Algebra. 

(a)  Monomials.     The  square   root  of  a  monomial   can 
usually  be  seen  by  inspection. 

36m%2=6  mzn  (because  (6  m2n)2  =36  m4n2).     Similarly, 


(b)  Trinomials.     If  a  trinomial  is  a  perfect  square,  its 
square  root  can  be  obtained  by  inspection. 

Thus  suppose  we  wish  to  find  the  square  root  of  9  z2  +  12  xy  +4  y*. 
This  trinomial  is  a  perfect  square  because  its  terms  9  z2  and  4  y* 
are  squares  and  positive,  while  its  remaining  term,  12  xy,  is  equal  to 
2-  A/9^2-  V41/2.  (See  §11  (d),  p.  20.)  Hence  the  trinomial 
can  be  expressed  in  the  form  (3  x  +2  yY,  whence  the  desired  square 
root  of  9  x2  +  12  xy  +4  yz  is  3  x  +2  y. 

Similarly,  V4s2-4s+l  =2s-l  because  4s2-4s+l  is  a  perfect 
trinomial  square,  and  as  such  is  factorable  into 

(2s-l)(2s-l)   or  (2s-l)2. 

(c)  Polynomials.     To  find  the  square  root  of  a  polynomial 
of  more  than  three  terms  we  may  follow  a  process  much  like 
that  employed  for  finding  square  roots  in  arithmetic.     This 
is  illustrated  in  the  following  example. 


VIII,  §  38]  SQUARE    ROOT  63 

EXAMPLE.  Find  the  square  root  of  4x4-f-12x3-3x2-18x+9. 

PROCESS. 

4x4+12x3-3x2-18x+9|  2x2+3x-3 
4  x4  Ans. 


Trial  divisor,          4  x2 
Complete  divisor,  4x2+3  x 


12x3-3x2 
12  x*+9  x2 


Trial  divisor,          4  x2-j-6  x 
Complete  divisor,  4x2+6  x  —  3 


-12x2-18x+9 
-12x2-18x+9 


EXPLANATION.  Arrange  the  terms  of  the  polynomial  in  the 
descending  (or  ascending)  powers  of  some  letter.  In  the  example, 
the  arrangement  is  in  descending  powers  of  x. 

Extract  the  square  root  of  the  first  term,  write  the  result  as  the 
first  term  of  the  root  (giving  the  2  x2  in  the  answer),  and  subtract 
its  square  from  the  given  polynomial  (giving  the  12  x3  —3  x2  in  the 
second  line  of  the  process). 

Divide  the  first  term  of  the  remainder  by  twice  the  root  already 
found,  used  as  a  trial  divisor.  The  quotient  (3  x}  is  the  next  term 
of  the  desired  root.  Write  this  term  in  the  root,  and  annex  it  to  the 
trial  divisor  to  form  the  complete  divisor  (the  4  x2  +3  x  of  the 
process). 

Multiply  the  complete  divisor  by  this  term  of  the  root,  and  subtract 
the  product  from  the  first  remainder  (giving  the  — 12  x2  — 18  x  +9 
of  the  process). 

Find  the  next  term  of  the  root  by  dividing  the  first  term  of  the 
remainder  by  the  first  term  of  the  new  trial  divisor.  This  gives  the 
—3  of  the  answer. 

Form  the  second  complete  divisor  and  continue  in  the  manner 
above  indicated  until  a  remainder  of  0  is  obtained. 

In  the  example  just  considered,  only  one  letter,  as  x, 
appears.  A  similar  process  may  be  employed,  however,  in 
all  cases  by  first  arranging  the  expression  in  descending  (or 
ascending)  powers  of  some  one  of  the  letters. 

For  example,  4  x4  +9  y*  — 12  x^y3  + 16  x2  + 16  -  24  yz,  when  arranged 
in  descending  powers  of  x,  becomes 


64  SECOND   COURSE   IN  ALGEBRA       [VIII,  §  38 

EXERCISES 

Find  (by  inspection  or  by  the  process  shown  in  §  38)  the 
square  root  of  each  of  the  following  expressions.  Check 
each  answer  by  squaring  it  to  see  if  the  result  thus  obtained 
is  the  given  expression. 

1.  4  xY-          4.   81  a866c10.     7.    196  ploq12.      10.   a?x2m. 

2.  9a664.  5.   225  ra8n4.      8.   m2nVr12.       11.   9  ra2r*4p. 

3.  25z22/4-36.      6.    625  rW.      9.    529  r10s14.      12.   m2*^. 

13.  z2+2z+l.  18.    9ra2-6raz-fz2. 

14.  z2-4z+4.  19.   x2+xy+±  y2. 

15.  4m2+12mn+9n2.  20.    9  z2+66  z+121. 

16.  4x2+4xy+y2.  21. 

17.  c2-4ac+4a2.  22. 

23.  4 

24.  z6 

25.  a;4 

[HINT.     See  remark  at  the  close  of  §  38.] 

26.  z8+2a6z2-a4z4-2a2z6+a8. 

[HINT.     First  arrange  in  descending  powers  of  x.] 

27.  9 

28.  9 

29.  z8+4  x7-3  x4-20  x5-2  x6+4+4  x2-  16  x+32  x*. 


39.  The  Double  Sign  of  the  Square  Root.  We  know  that 
3  is  the  square  root  of  9  because  32  =  9.  But  we  also  have 
(  —  3)2  =  9.  Therefore,  —3  can  also  be  regarded  as  a  square 
root  of  9.  In  other  words,  9  has  two  square  roots,  +3  and 
—  3,  which  are  opposite  in  sign  but  otherwise  the  same. 
Similarly,  16  has  the  two  square  roots  +4  and  —4,  and  in 
general,  a2  has  the  two  roots  a  and  —  a. 

The  double  sign  ±  is  sometimes  used.  Thus  we  say  that  the 
square  root  of  9  is  ±3.  This  is  merely  a  brief  way  of  saying  that  the 
two  roots  are  +3  and  —3. 


VIII,  §40]  SQUARE    ROOT  65 

In  order  to  avoid  all  confusion,  it  is  to  be  understood  here- 
after that  the  radical  sign  V~  when  placed  over  a  number 
means  the  positive  square  root  of  that  number.  If  it  is  de- 
sired to  indicate  the  negative  square  root,  it  is  done  by  the 
symbol  —  V  . 

Thus  Vl6  means  +4,  while  -  Vl6  means  -4.  Similarly,  Va 
means  +Va. 

40.  Equations  Containing  Radical  Signs.  Equations  con- 
taining radical  signs  may  often  be  solved  by  squaring  each 
member.  This  is  equivalent  to  multiplying  each  member  by 
the  same  amount,  and  hence  is  justified  by  §  27. 

EXAMPLE  1.     Solve  the  equation  Vx  —  2  =  6. 
SOLUTION.     Squaring  both  members  gives 

z-2=36, 

whence  x=38.     Ans. 

CHECK.  V38-2  =  V36~=6. 

EXAMPLE  2.     Solve  the  equation  Vz— 1  — VE— 4  =  1. 


SOLUTION.     Transpose  the  Vx  -4  to  the  right  ;   this  gives 


Square  both  members,  using  Formula  VI,  §  10  for  finding 
(1  +  Vc-4)2.     This  gives 


or 


Canceling  x  from  both  sides  and  transposing  the  1  and  —4  to 
the  left,  gives  _ 

2  =2  Vc-4,  or  Vz-4  =  l, 
whence  (squaring  again) 

z-4  =  !2  =  l. 

Therefore  x=5.     Ans. 

CHECK.        V5^1  -  V5^4  =  V4-  VI  =2-1  =1. 


66  SECOND   COURSE    IN   ALGEBRA       [VIII,  §  40 

NOTE.  It  is  especially  important  to  check  all  the  answers  ob- 
tained for  equations  containing  radical  signs,  since  the  process  of 
squaring  both  members  sometimes  leads  to  a  new  equation  whose 
roots  do  not  all  belong  to  the  first  one.  Thus,  if  we  square  both 
members  of  the  equation  x  =5,  we  get  x2  =25  and  this  last  equation 
has  —5  as  a  root  as  well  as  5. 

EXERCISES 

Solve  each  of  the   following   equations  and  check  each 


answer. 


3. 


4.    Vx+7-Vx  = 


[HINT.     First  transpose  one  of  the  radicals  to   the  right  side, 
as  in  Example  2,  §  40.] 


9. 


10.  If  16  be  added  to  4  times  a  certain  number,  the  square 
root  of  the  result  is  6.     What  is  the  number? 

11.  If  9  be  added  to  the  square  of  a  certain  number,  the 
square  root  of  the  result  is  5.     What  is  the  number? 

12.  The  difference  between  the  square  root  of  a  certain 
number  and  the  square  root  of  11  less  than  that  number  is  1. 
Find  the  number. 

13.  Solve  each  of  the  following  equations. 
(a) 

(6) 


CHAPTER  IX 
RADICALS 

41.  Radicals.  Suppose  we  have  a  square  which  we  know 
contains  exactly  2  square  feet.  How  long  is  each  of  its  four 
sides?  In  order  to  answer  this,  we  naturally  let  x  represent 
the  desired  length.  Then  we  must  have 


x  -x  =      or 


2  sq.ft. 


Therefore         x  =  A/2  ft.     Ans. 

This  number   A/2    cannot  be   determined 
exactly  because   it  is  the  square  root  of  a          pIG  10 
number,  2,   which  is  not  a  perfect  square. 
However,  A/2  measures  a  perfectly  definite  length,  as  indi- 
cated in  the  figure.     Its  value,  correct  to  two  decimal  places 
only,  is  1.41. 

Such  a  number  as  A/2  is  called  a  quadratic  radical.  This 
name  is  used  in  general  to  denote  the  indicated  square  root 
of  a  number. 


Thus   V3,  V7,  Vzi,  VI06,  V213  are  all  radicals. 

The  word  radical  is  also  used  in  connection  with  other  roots 
than  square  roots.  Thus  VlO  means  the  cube  root  of  10, 
that  is  the  number  whose  cube  is  10.  Similarly,  v^6  means 
the  fourth  root  of  6,  etc.  All  such  numbers  represent  per- 
fectly definite  magnitudes,  as  did  \/2  in  the  figure  above, 
even  though  we  cannot  express  them  exactly  in  decimal  form. 

67 


68  SECOND   COURSE   IN  ALGEBRA          [IX,  §  41 

In  general,  the  nth  root  of  any  number  a  is  written  A/a,  and 
this  is  known  as  a  radical  of  the  nth  order.  The  number  n  is 
here  called  the  index  of  the  root,  and  the  number  a  is  called 
the  radicand. 

NOTE.  When  no  index  is  expressed,  the  index  2  is  to  be  under- 
stood. Thus  V3  means  \/3. 

The  same  definitions  apply  also  to  algebraic  expressions. 
Thus  V3xy*  and  Vat+b2  are  radicals. 

42.  Rational  and  Irrational  Numbers.  Surds.  The  posi- 
tive and  negative  integers  and  fractions,  and  zero,  are  called 
rational  numbers.  If  an  indicated  root  of  a  number  cannot 
be  extracted  exactly,  that  is,  cannot  be  expressed  exactly  as 
one  of  these  rational  numbers,  it  is  called  a  sure?.  Any  posi- 
tive or  negative  number  that  is  not  rational  is  called  irrational 

Thus_  V3,  ^10,  V6  are  surds ;  but  V§,  v^S,  Vj^  are  rational, 
since  Vg=3,  ^8=2, 


EXERCISES 

Determine  which  of  the  following  radicals  are  surds ;   and 
state  the  index  and  the  radicand  of  each. 

1.  V7.     2.  V8.     3.  Vl6".     4.  V^T     5.  Vff .     6.  V75. 
7.  V^S. 
[HINT.  -8  =  (-2)3.] 

8.  Vis.      10.  Vs.      12.  V20.      14.  Vs^y. 

9.  V27.         11.  Vl6.       13.  V32.         15.  Vg  m6(a+6)9. 


43.  Value  of  Radicals.  Use  of  Table.  To  determine  the 
value  of  a  radical  correct  to  two  or  more  decimal  places 
usually  calls  for  a  rather  long  process.  (See  §  37,  p.  61.) 
In  order  to  save  time  and  labor,  the  values  of  those  radicals 
which  are  needed  most  in  ordinary  life  (the  square  and  cube 


IX,  §43]  RADICALS  69 

roots)  have  been  printed  in  a  table  and  placed  for  convenient 
reference  at  the  end  of  this  book.  For  the  sake  of  complete- 
ness, the  second  and  third  powers  of  numbers  are  also  printed 
in  the  table.  Just  how  to  use  this  table  is  described  on  page 
275,  which  the  pupil  should  now  read  carefully.  Below  are 
a  few  illustrative  examples. 

EXAMPLE  1.     Find  \/7,  using  the  table. 

SOLUTION.  The  top  number  in  the  third  column  on  page  290 
(table)  gives  V?  =2.64575.  This  value  is  correct  up  to  the  last 
decimal  figure  given,  that  is  to  the  fifth  place.  Thus  the  answer 
may  be  written  "^  =  2.64575+,  the  sign  +  indicating  that  this  value 
for  V7  is  correct  up  to  the  last  decimal  place  stated. 

EXAMPLE  2.     Find  vT,  using  the  table. 
SOLUTION.     The  top  number  in  the  sixth  column  on  page  290 
(table)  gives  ^7  =  1.91293+.     Ans. 

EXAMPLE  3.     Find  A/70. 

SOLUTION.  The  top  number  in  the  fourth  column  on  page  290 
(table)  gives  V70  =8.36660+.  Ans. 

EXAMPLE  4.     Find  v/70. 

SOLUTION.     V70  =  4.12129+,  from  the  seventh  column,  page  290. 

EXAMPLE  5.     Find  ^700. 

SOLUTION.     v'TOO  =8.87904+,  from  the  eighth  column,  page  290. 

EXERCISES 

By  means  of  the  table,  determine  the  approximate  values 
of  the  following  radicals. 

1.    V6.       2.  V60.        3.    ^6.       4.    v"60.       5.    ^500. 

6.  VeT. 

[HINT.  61=6.10X10.  So  use  the  information  given  for  6.10 
in  the  table.] 


70  SECOND    COURSE   IN   ALGEBRA 

7.    S/6L      8.    V92".       9.  -v/920. 

12.    \/78^2        13.    A/782. 

SOLUTION.  782  =  7.82  X  100.  Therefore  V782  is  the  same  as 
V7.82  except  that  the  decimal  point  in  the  root  must  be  moved 
one  place  farther  to  the  right.  (See  p.  275.)  Now,  V7.82  =  2.79643 
(table),  so  V782  =27.9643.  Ans. 


14.  A/561. 

[HINT.     See  Solution  of  Ex.  13.] 

15.  A/779.  16.    A/895". 

17.    ^6120. 

[HINT.     6120  =6.12  XlOOO.     (See  p.  276.)] 

18. 


19. 

[HINT.     .67  =^  X6.7     (Now  see  p.  276.)] 

20.    V.0676 

APPLIED   PROBLEMS 

Use  the  tables  in  working  the  following  problems. 

1.  If  the  sides  of  a  right  triangle  are  3  inches  and  2  inches 
long,  respectively,  what  is  the  length  of  the  hypotenuse? 

[HINT.     If  x  be  the  hypotenuse,  then  x2  =32+22  =  13.] 

2.  A  baseball  diamond  is  a  square  90  feet  on  a  side.     How 
far  is  it  from  home  plate  to  second  base  ? 

3.  If  the  diagonal  of  a  square  is  13  feet  long,  how  long  is 
each  side? 

4.  The  dimensions  of  a  certain  rectangular  field  are  103 
feet  by  337  feet.     In  going  from  one  corner  to  the  opposite 
corner,  how  much  shorter  is  it  to  go  by  the  diagonal  than  to 
go  around? 


IX,  §441  RADICALS  71 

5.    How  long  must  the  radius  of  a  circle  be  in  order 
that  the  area  be  12  square  inches?     (See  Ex.  14  (6),  p.  6. 


6.  What  is  the  length  of  the  edge  of  a  cube  if  the  volume 
is  357  cubic  inches? 

7.  If  the  volume  of  a  sphere  is  440  cubic  inches,  how  long 
is  its  radius  ?     (See  Ex.  14  (e),  p.  6.) 

8.  In  the  accompanying  figure  how 
long  should  the  radius  of  the  inner  semi- 

circle be  in  order  that  the  area  inclosed    ^  g  •  -  Q  D 

may  be  132  square  feet?  AB  =  CD=  2  feet 

FIG.  11. 

9.  The  area  A  of  a  triangle  in  terms 

of  its  three  sides,  a,  b,  and  c,  is  A  =  Vs  (s  —  a)  (s  —  b)  (s  —  c)  , 

where  s  =  —     The  sides  of  a  triangle  are  respectively 

2i 

6  inches,  7  inches,  and  9  inches  long.     What  is  the  area  ? 

10.  Two  circular  cones  have  altitudes,  h,  which  are  the 
same,  but  their  bases  have  different  radii.  What  is  the 
ratio  of  the  longer  radius  to  the  shorter  if  the  volume  of  the 
one  cone  is  three  times  that  of  the  other? 

[HINT.  The  formula  for  the  volume  of  a  cone  is  V  =  £irr2/&,  where 
h  represents  the  altitude  and  r  is  the  radius  of  the  base.] 

44.  Simplification  of  Radicals.  We  know  that  the  square 
root  of  the  product  of  two  numbers  is  the  same  as  the  product 
of  their  square  roots.  For  example,  V4x25  is  the  same  as 
A/4  X  V25,  because  both  are  equal  to  10  (the  first  being  VIM, 
or  10,  and  the  second  being  2  X5,  or  10).  In  the  same  way, 
v/8x3  =  v/8XV/3,  or  simply  2^3.  In  fact,  we  have  the 
following  general  formula. 


72  SECOND   COURSE   IN   ALGEBRA          [IX,  §44 

Formula  I.  Vafc  =  Va  •  V&. 


Again,  Vf  is  the  same  as  —  -  because  both  are  equal  to  f . 

V9 

(Explain.)     Similarly,  v^f  may  be  written  -— ,  or  — —     So 

V  8          2 
in  general  we  have  the  following  formula. 

Formula  II.  V1/? 


Formulas  I  and  II  enable  us  to  simplify  many  radical  ex- 
pressions, as  is  illustrated  in  the  following  examples. 

EXAMPLE  1.     Simplify  V63. 
SOLUTION.     Using  Formula  I,  we  have 

V63  =  V9X7  =  V9  X  V?  =3  V?.     Arcs. 


EXAMPLE  2.     Simplify  \/32. 

SOLUTION.     ^32=4/8X4=^/8X^4=2^4.     Ans. 

EXAMPLE  3.     Simplify 

/8       Vg      V4x2     Vix^/2     2\/2 
SOLUTION.     \~  =  — — :  =        ^    =  — — -  — —  =  — — • 

^27     V27     \/9x3    .V9XV5    3V3 

EXAMPLE  4.     Simplify  V20  a6". 

SOLUTION.     V20  a6  =  V4a6x5  =  V4^ x  V5  =2  a3  Vs.     Ans. 


EXAMPLE  5.     Sunplify  v/ 


SOLUTION. 


)  X  (9  x2)  =  v'S 

*         Z*  ^6  22  22 

NOTE.     It  will  be  observed  that  in  each  of  the  examples  above 
the  process  of  simplification  consists  in  removing  from  under  the 


IX,  §  44]  RADICALS  73 

radical  sign  the  largest  factor  of  the  radicand  which  is  a  perfect 
square,  or  perfect  cube,  as  the  case  may  be.  Thus,  in  Example  1, 
the  radicand,  63,  was  first  broken  up  into  factors  in  such  a  way  that 
9  (which  is  a  perfect  square)  appears  clearly  to  the  eye.  Similarly, 
in  Example  2  (where  we  are  dealing  with  a  cube  root)  we  first  write 
the  radicand,  32,  in  a  form  which  brings  out  conspicuously  its 
factor  8,  which  is  a  perfect  cube.  The  first  step  in  all  such  examples 
is,  therefore,  to  get  the  radicand  properly  broken  up  into  factors. 
This  requires  good  judgment,  but  becomes  very  easy  after  slight 
practice  and  experience. 

EXERCISES 

1.  By  Formula  I,  p.  72,  we  have  A/20  =  A/4x5  =  A/4  X  A/5 
=  2A/5.     Look  up  the  values  of  A/20  and  A/5  in  the  table 
and  thus  prove  that  A/20  is  the  same  as  2 A/5. 

2.  Show  that  Formula  I,  p.  72,  gives  A/54  =  3A/(T  and 
verify  the  correctness  of  this  result  by  use  of  the  table,  as  in 
Ex.  1. 

Simplify  each  of  the  following  radicals.     (See  Note  in  §  44.) 

3.  A/18.  6.    A/125.  9.    v/54. 

4.  A/24.  7.    A/108.  10.    A^Sl. 
6.    A/Tl2i             8.    A^32. 

11.    V||. 

[HINT.     First  use  Formula  II,  §  44.] 

12-   ^5S-  19.    A^27  x*y*z\ 

20.    \/16  M4. 

o3/ 

6  a?bVab.  Ans. 


4(a2-62) 


74  SECOND   COURSE    IN   ALGEBRA          [IX,  §  44 

Write  each  of  the  following  in  a  form  having  no  coefficient 
outside  the  radical  sign. 

23.   2>/3. 

SOLUTION.  By  Formula  I,  §  44, 2^3  =  Vix  \/3  =  Vl2.     Ans. 
24.    2V2".  25.    5V5^  26.    2v^ 

27.   W2. 

A/9         A/9 

[HINT.     Write  W2  =-rr  =-^,  and  apply  Formula  II,  §  44.] 
3       Vg 

33.  2gVx-y.         38.  fV3. 
.  Ans.     34.  ra^Xn.  39.  f^S. 

35.  2aV«2-62.       40.  fVf. 

36.  ^L-  41.  (a-6)V2c". 


42.    o- 


32.  SmnVmn.  37.  r\'-- 


45.  Addition  and  Subtraction  of  Similar  Radicals.  When- 
ever two  radicals  having  the  same  index  have  also  the  same 
radicand  (or  can  be  given  the  same  radicand  by  simplifica- 
tion) they  are  called  similar  radicals. 

Thus  2  V2  and  3  V2  are  similar  radicals  ;  so  also  are  V2  and  V32 
since  the  last  of  these  may  be  simplified  into  4V2,  by  §  44.  Like- 
wise, V3  a2x  and  V3  b2x  are  similar,  being  equal,  respectively,  to 
a  VWx  and  6  V3  x,  thus  coming  to  have  the  same  radicand. 

Whenever  similar  radicals  are  added  or  subtracted,  the 
result  may  be  expressed  in  a  single  term. 

Thus  4V3+5V3  =  (4+5)\/3=9\/3.     Ans.  _ 

Again,          3V32-2V8=3X4V2-2X2V2~=12V2-4V2 
=  (12-4)V2=8V2.     Ans. 

Likewise, 

2 V4~cM>  +  V9  azb  -  Vl6  a*b  =2  •  2  avT+3  a*N/6_-4  aV6 

=  (4  a+3  a-4  a)V6=3  aVF.     Ans. 


IX,  §46] 


RADICALS 


75 


NOTE.  The  pupil  is  especially  warned  that  in  general  we  cannot 
write  Va+6  =  Va  +  Vb.  Thus  when  a  =4,  6=9  this  would  give 
Vl3=2+3=5,  which  is  clearly  false.  Similarly,  we  cannot  write 
V~a~—b  =  Va-Vb.  Thus V25^9=Vl6=4, but V25-V9=5-3  =  2. 


EXERCISES 

Combine  the  following  radicals. 

1.  V8+V18+V32.  Check  your  answer  by  use  of 
the  table;  that  is  show  that  \/8+ VT8+V32,  as  computed 
by  the  table,  has  the  same  value  as  your  answer,  similarly 
computed. 

2.  V108+V27-V75. 

3.  v'm-h^lG-v^. 

4.  V72+V32-V50". 


Check  your  answer  as  in  Ex.1. 
Check  as  in  Exs.  1  and  2. 

7.       32  a2- 

8. 

9. 


46.  Multiplication  of  Radicals.  We  may  multiply  one 
radical,  as  Va,  by  another  of  the  same  index,  as  \/b,  by  For- 
mula I  of  §  44.  If  n  —  2,  this  gives  as  an  important  special 
case 

Va  •  Vb  = 


76  SECOND   COURSE    IN   ALGEBRA          [IX,  §46 

EXAMPLE  1.     Find  the  product  of  V2^and  Vl8. 
SOLUTION.     V%  >  Vl8  =  V2  •  18  =  V36  =6.     Ans. 

EXAMPLE  2.     Multiply  V3+A/5  by  2  A/3-  \/5. 
SOLUTION.  V3+   V5 

2V3-   V5 

2-  3+2VI5 
-   VI5-5 


6+         5-5  =  l  +  Vl5.     Ans. 

EXAMPLE  3  .     Multiply  Va  +  Va  —  b  by  Va  —  Va  —  b. 
SOLUTION.      Va  +  Va  —  b 
Va-  Va-b 
a  +  Va2—ab 

—  Va2—  ab  —  (o-b) 
a  —  (a  —  6)  =a—  a+b  =  b.     Ans. 

EXERCISES 

Find  the  following  products,  simplifying  results  as  far  as 
possible. 

1.  V3  -  \/27.  7.  V7  .  \/I 

2.  V8  -  Vl2.  8.  VI  -  VTOl. 

3.  V6-V4.  9.  (V3-V^)(\/3+>/2). 

4.  \/7  •  V9.  10.  (2v/3-V2)(2\/3+v/2). 

5.  VI6  -  V3  -  V2.  11.  (V6-  V3)2. 

6.  \/f  .  V|.  12. 

13.  (3\/3+2V5)(V3-3\/5). 

14.  (V2+V3  +  V5)(V2-V3). 

15.  Va-.Vtf.  18. 

16.  Vo6  •  V^3.  19.    (Va-\/6)2. 

17.  Vtf-V-  20. 


IX,  §47 


RADICALS 


77 


21. 

22.   Find  the  value  ofz2-4z-lifa:  =  2+  \/5. 


23.   Find  the  value  of  z2+3z-2if 


24.  Does  A/3+2  satisfy  the  equation  x?  —  4  z+1  =  0 ;  that 
is,  is  the  equation  true  when  x  =  A/3+2  ?  Answer  the  same 
question  when  x  =  A/3 + A/2. 

47.  Division  of  Radicals.  We  may  divide  one  radical,  as 
Va,  by  another  of  the  same  index,  as  V6,  by  the  Formula  II 
of  §  44.  If  n  =  2,  this  gives  as  an  important  special  case 

*-4 


EXERCISES 

Express  each  of  the  following  quotients  as  a  fraction  under 
one  radical  sign,  and  reduce  your  answer  to  simplest  form. 

A/15 


Ans. 


10.    V81+9  a2+a4^- Va2-3  a+9. 


CHAPTER  X 
QUADRATIC   EQUATIONS 

48.  Quadratic  Equation.     An   equation  which   contains 
the  unknown  letter  to  the  second  (but  no  higher)  power  is 
called  a  quadratic  equation,  or  briefly,  a  quadratic. 

Thus  the  equations  2  x2  —  4  x  =  I  and  ^  x2  +x  —  —  3  are  quadratics, 
but  2  x  -3  =0  and  4  x3  -5  x2+x  =2  are  not. 

49.  Pure   Quadratic.     When  the  quadratic  contains  the 
second  power  only  of  the  unknown  letter,  it  is  called  a  pure 
quadratic. 

Thus  2  x2  —27  =0  and  ax"2  =  be  are  pure  quadratics,  but  x2  —  4  x  =2 
and  x2+bx+c=Q  are  not. 

50.  Affected   Quadratic.     When  the  quadratic  contains 
both  the  first  and  second  powers  of  the  unknown  letter,  it  is 
called  an  affected  quadratic. 

Thus  z2+3z=7  and  xz-\-2ax=az  are  affected  quadratics,  but 
2  x2  -7  =0  and  5  x2  -16  o262  =c2  are  not. 

51.  Solution  of  Pure  Quadratics.     The  following  example 
will  suffice  to  show  how  the  solution  of  any  pure  quadratic 
may  be  obtained. 

EXAMPLE  .     Solve  2  x2  -  30  =  0. 

SOLUTION.     Transposing  and  dividing  through  by  2  gives^2  =  15. 
Taking  the  square  root  of  both  members  gives  x  =  =  Vl5.    Ans. 
To  get  the  approximate  value  of  Vl5,  we  may  consult  the  table, 
where  we  find  Vl5  =3.87298+. 

The  answer  may,  therefore,  be  written  in  the  form  x  =  ±3.87298+. 
CHECK.     2 (Vl5)2- 30  =2X15 -30  =30 -30=0,  as  required. 
2(  -  Vl5)2 -30  =2 X 15  -30  =30 -30  =0,  as  required. 
78 


X,  §  51]  QUADRATIC   EQUATIONS  79 

NOTE.  Strictly  speaking,  when  we  extract  the  square  root  of 
both  members  of  the  equation  x2  =  15  we  get  ±3  =  ±  VlS.  But  to 
say  that  —  x  =  =«=  Vl5  means  the  same  as  +x  =  =±=  Vl5,  so  it  suffices 
to  write  simply  x  =  =±=  Vl5  to  cover  all  cases. 

An  examination  of  the  example  above  shows  that  we  have 
the  following  rule. 

To  solve  a  pure  quadratic,  solve  for  x2,  then  take  the  square 
root  of  the  result.  There  will  be  two  solutions,  the  one  being 
the  negative  of  the  other. 

EXERCISES 

Solve  each  of  the  following  equations,  checking  your 
answer  for  the  first  five.  If  you  meet  with  a  radical,  find  its 
approximate  value  by  use  of  the  table. 

1.  z2-81  =  0.  11       3    _1  j  •     1 

2.  3*2-192  =  0.  '   *+5          2*~5' 

12     JL         !     _Z_L    1 
'  ' 


3.  4z2+8  =  10*2-16. 

4.  3x2-15  =  0.  x+3    ,2x-l=Q 

~ 


5.  3,2-!6  =  0. 

„  _0 

6.  3z2-17  =  0.  -'     —- 

?     ^_^    =19 


x-2  x+3 

[HINT.     See  §  40.] 


17. 


10.    (o;+l)2-2(x+l)=4.       18.    V25-6  x+ V25+6  a;  =  8. 


80  SECOND   COURSE    IN   ALGEBRA  [X,  §  51 

APPLIED   PROBLEMS 

1.  What  numbers  are  equal  to  their  own  reciprocals? 

2.  One  side  of  a  right  triangle  measures  3  inches  and  the 
hypotenuse  measures  7  inches.     Find  (approximately)  the 
length  of  the  other  side. 

[HINT.  Work  by  algebra,  making  use  of  the  principle  that  in 
any  right  triangle  the  square  of  the  hypotenuse  equals  the  sum 
of  the  squares  of  the  two  sides.] 

3.  What  is  the  length  of  the  longest  umbrella  than  can 
be  placed  in  the  bottom  of  a  trunk  the  inside  of  which  is  33 
inches  long  by  21  inches  wide? 

4.  A  certain  square  has  a  side  which  is  three  times  as  long 
as  the  side  of  another  square.     If  the  difference  of  their  areas 
is  72  square  feet,  how  long  is  the  side  of  each  ? 

5.  Find  the  mean  proportional  between  25  and  9;    also 
that  between  17  and  21.     In  what  particular  is  the  latter  one 
essentially  different  from  the  first  one  ? 

6.  It   is   proved  in  geometry  that  whenever  a  perpen- 
dicular is  drawn  from  a  point  on  a  semicircle  to  the  base, 
as  PQ  in  Fig.  12,  its  length  is  a  mean  Q 
proportional  between  the  segments  AP 

and  PC  of  the  base ;  that  is, 
AP  =  PQ 
PQ     PC 

IfAP  =  8  inches  and  PC  =  10  inches, 
how  long  is  PQ  ? 

7.  Determine  the  formula  for 

(a)  The  side  of  the  square  whose  area  is  a. 

(b)  The  radius  of  the  circle  whose  area  is  a. 


X,  §  52]  QUADRATIC   EQUATIONS  81 

(c)  The  radius  of  the  sphere  whose  area  is  a. 
[HINT.     See  Ex.  14,  p.  6.] 

(d)  The  diameter  of  the  base  of  a  circular  cone  whose 
volume  is  v  and  whose  altitude  is  h. 

(HINT.  The  volume  of  a  circular  cone  is  equal  to  the  area  of  its 
base  multiplied  by  -^  its  altitude,  i.e.,  V  =  ^Trr2h.] 

8.  The  distance  s,  measured  in  feet,  through  which  an 
object  falls  in  t  seconds  when  dropped  vertically  downward  is 
given  by  the  formula  s  =  %gt*,  where  0  =  32  (approximately). 
Hence,  determine  (approximately)  how  long  it  will  take  a 
stone  to  drop  to  the  bottom  of  a  mine  300  feet  deep. 

9.  One  of  the  sides  of  a  certain  triangle  is  m  units  long. 
What  is  the  formula  for  the  corresponding  side  of  a  similar 
triangle  whose  area  is  n  times  as  great? 

[HINT.  It  is  shown  in  geometry  that  if  two  geometric  figures  are 
similar ;  that  is,  have  the  same  shape  but  are  of  different  sizes,  then 
the  square  of  any  line  in  the  first  figure  is  to  the  square  of  the  cor- 
responding line  in  the  second  figure  as  the  area  of  the  first  figure  is 
to  the  area  of  the  second.] 

10.  A  map  of  the  United  States  is  uniformly  enlarged  in 
such  a  way  as  to  cover  twice  as  much  area  on  the  paper  as 
before.     By  what  factor  should  the  scale  of  the  map  be  now 
multiplied? 

11.  Find  three  consecutive  integers  such  that  the  square 
of  the  second  plus  the  product  of  the  other  two  equals  31. 

[HINT.  Let  x  be  the  second  integer.  Then  the  first  and  last 
integers  will  be  x  — 1  and  x  +  1  respectively.] 

52.  Solution  of  Affected  Quadratics  by  Factoring.  It  is  a 
familiar  principle  of  arithmetic  that  the  product  of  two 
numbers  is  zero  if  either  of  the  numbers  is  zero ;  that  is,  if 
either  factor  is  zero. 

For  example  2X0=0,  0X4=0,   (-3)  X0=0,  etc. 
G 


82  SECOND   COURSE    IN   ALGEBRA  [X,  §  52 

This  principle  is  frequently  used  to  solve  affected  quadratic 
equations. 

EXAMPLE.     Solve  by  factoring  the  equation  #2+8  x  =  48. 
SOLUTION.     Transposing  all  terms  to  the  left,  we  have 


Factoring, 

z2+8z-48  =  (z-4)(z  +  12).         [§  11  (c)] 

Thus  the  given  equation  becomes 


This  equation  will  be  satisfied,  according  to  the  principle  men- 
tioned above,  whenever  the  factor  x—  4  equals  zero  or  the  factor 
z  +  12  equals  zero,  that  is  in  case  x—  4=0  or  a;  +  12=0.  Solving 
these  two  simple  equations  gives  z=4  and  x=  —  12,  which  must 
therefore  be  the  desired  solutions. 

CHECK.     When  x  =  4  the  left  side  of  the  original  equation  becomes 
42+8x4,  which  reduces  to  16+32  =48,  as  the  equation  demands. 
When  x  =  —  12  we  have  in  like  manner 

(-12)2+8x(-12)  =  144-96=48. 

We  thus  have  the  following  rule. 

To  solve  quadratics  by  factoring  : 

1.  Transpose  all  terms  to  the  left  so  as  to  have  0  on  the  right. 

2.  Factor  the  left  member  of  the  resulting  equation. 

3.  Place  each  factor  equal  to  0  and  solve  the  resulting  simple 
equations.     The  two  results  are  the  solutions  required. 

EXERCISES 

Solve  each  of  the  following  equations  by  factoring,  checking 
your  answer  in  the  first  five. 

1.  z2-7z+10  =  0.  3.   z2+8z=-15. 

2.  z2-5z=-6.  4.   z2+7z-30  =  0. 


X,  §53]  QUADRATIC    EQUATIONS  83 

5.  z2 

6.  l- 
7. 


[HINT.     Write  as 
8.    x2-l  =  3 


5  then  apply  the  principle  in  §  52.] 

~z+4  =  13.   3(z+l)(z-3)+4(z-3)=0. 

10.    j£-  =  __  i  __  5.        14-    (z+l)2+3(z+l)+2  =  0. 
x  —  2         x  —  2  [HINT.     Solve  first  for  x+L] 

53.  Solution  of  Any  Quadratic  by  Completing  the  Square. 
We  often  meet  with  a  quadratic,  such  as  xz-\-7  x  —  5  =  0, 
which  we  cannot  solve  as  in  §  52  by  factoring.  The  difficulty 
here  is  that  we  cannot  factor  readily  x2+7  x  —  5.  However, 
this  quadratic  and  all  others  (whether  solvable  by  factoring 
or  not)  can  be  solved  by  a  certain  process  known  as  completing 
the  square.  How  this  is  done  will  be  best  understood  from  a 
careful  study  of  the  following  examples. 

EXAMPLE  1.     Solve  z2+  6  x  =  16. 

SOLUTION.  The  first  member  of  this  equation,  or  x2  +6  x,  would 
become'  a  trinomial  square  [§  ll(d)]  if  9  were  added  to  it.  Our 
first  step,  therefore,  is  to  add  9  to  both  members  of  the  given  equa- 
tion, thus  "completing  the  square"  in  the  first  member  and  giving 
us  the  equation  X2  _j_6  x  _jig  =  25, 

or  (x  +3)2=25. 

Taking  the  square  root  of  both  members  of  the  last  equation  is 
now  an  easy  process  and  gives 

x  +3  =  ±5. 

Therefore  we  must  either  have  #+3  =5,  or  x+3  =  —5. 

Solving  the  last  two  equations  gives  as  the  desired  solutions 
x=2andx=—  8.  Ans. 

CHECK.  Substituting  2  for  x  in  the  first  member  of  the  given 
equation  gives  22+6X2,  which  reduces  to  4  +  12  =  16,  as  desired. 
Similarly,  with  x  equal  to  —8,  the  first  member  of  the  given  equa- 
tion becomes  (-8)2+6X(-8),  or  64—48,  which  reduces  to  16  as 
required. 


84  SECOND   COURSE   IN   ALGEBRA  [X,  §  53 

EXAMPLE  2.     Solve  x2-  8  x+  14  =  0. 

SOLUTION.     Transposing,  x2  —  8  x  =  —  14. 

Completing  the  square  by  adding  16  to  both  sides  gives 

z2-8z  +  16=2,  or  (z-4)2=2. 
Taking  the  square  root  of  both  members, 

x-4=±V2. 
Solving  the  last  two  equations, 

z=4  +  V2andz=4-  V2.    Ans. 

CHECK.  With  x=4  +  V2,  the  first  member  of  the  given  equation 
becomes  (4  +  V2)2-8(4  +  V2)  +14.  By  Formula  VI  of  §10  this 
may  be  written 

(16+8V2+2)-8(4  +  V2)+14,  or  16+8V2~+2-32-8V2  +  14. 

Here  the  8>/2  and  the  -8^2  cancel,  while  the  rest  of  the  expres- 
sion (namely  16+2—32  +  14)  reduces  to  0,  as  required. 

Likewise,  when  x  has  its  other  value,  namely  x  =4  —  V2,  the  first 
member  may  be  shown  to  become  0. 

NOTE.  Since  the  solutions  obtained  above  for  Example  2  con- 
tain the  surd  V2,  they  cannot  be  expressed  exactly  (see  §  37),  but 
we_can  express  their  values  approximately.  Thus,  the  table  gives 
V2  =  1.41421+  so  that  the  two  solutions  become  4  +  1.41421+  and 
4-1.41421+,  which  reduce  to  5.41421+  and  2.58579+.  Ans. 

EXAMPLE  3.     Solve  3  z2+8  x  =  15. 

SOLUTION.  Dividing  through  by  3  so  as  to  have  +1  as  the  co- 
efficient of  x2,  the  equation  becomes 

z2+fz=5. 
Completing  the  square  by  adding  (-|)2  (or  -^-)  to  both  sides  gives 

or, 


Taking  the  square  root  of  both  members, 

x  +  *  =  ± 

Therefore,  the  two  solutions  are 

z=-f+iV61  and  z 

These  two  answers  may  be  written  together  in  the  condensed 
form  x  =-^(  —  4±  V61)  and  by  looking  up  the  value  of  V61  in  the 
tables,  these  values  of  x  may  be  determined  approximately,  as  in- 
dicated in  the  Note  to  Example  2. 


X,  §  54]  QUADRATIC   EQUATIONS  85 

54.  Summary  and  Rule.  It  is  now  to  be  observed  care- 
fully that  in  each  of  the  three  examples  just  considered 
(§  53)  the  first  step  in  the  solution  consists  in  reducing  the 
given  equation  to  the  type  form 


where  p  and  q  are  given  numbers. 

Thus,  in  Example  3,  we  first  put  the  equation  3  x2  +8  x  =  15  into 
the  form  z2+f  z=5.  Here  p=f,  and  q  =5. 

The  next  step  is  to  complete  the  square.  This  is  done  in 
each  case  by  adding  to  both  members  the  square  of  half  the 
coefficient  of  x,  that  is  we  add  (p/2)2  to  both  members. 

Thus,  in  Example  3,  we  had  p=f,  so  we  added  (|-)2  to  both 
members. 

After  this,  the  equation  is  such  that  we  can  extract  the 
square  root  of  the  left  member,  and  when  we  do  so  and 
equate  results,  we  obtain  two  simple  equations,  each 
yielding  a  solution  of  the  given  quadratic. 

This  may  now  be  summarized  in  the  following  rule. 

To  solve  any  quadratic  : 

1.    Reduce  the  equation  to  the  form 


2.  Complete  the  square  by  adding  (p/2)2  to  both  members. 

3.  Extract  the  square  root  of  both  members  of  the  new  equa- 
tion and  equate  results.     This  yields  the  two  solutions  desired. 

EXERCISES 

Solve  each  of  the  following  equations,  checking  your  answer 
in  the  first  five. 

1.  z2-5z  =  14.  6.  8z  =  z2-180. 

2.  z2-20z  =  21.  7.  z2+22z=-120. 

3.  z2-12z+20  =  0/  8.  2/2  =  10-3i/. 

4.  x2-2x  =  ll.  9.  z2- 
6.   z2-3z-5  =  0.  10.  6z2 


86  SECOND   COURSE    IN   ALGEBRA  [X,  §  54 

11.    2z2+ 


12.  1-3  *  =  2*2. 

17.  ^— 

13.  2  £(z-f-4)=42.  x 

14.  (3x-2)2  =  6z+ll,  18.  -^ 


15.   x+—  =  16. 


. 

z+5     z-2 

55.     Solution  of  Quadratics  by  the  Hindu  Method.     A 

simple  way  preferred  by  many  for  completing  the  square  in 
any  quadratic  is  the  one  called  the  Hindu  method.  It  con- 
sists of  two  steps  : 

1.  Multiply  both  members  by  four  times  the  coefficient  of  x2. 

2.  Add  to  both  members  of  the  new  equation  the  square  of 
the  original  coefficient  of  x. 

EXAMPLE.     Solve  2  x2  -  3  x  =  2. 

SOLUTION.     Multiplying  through  by  4  times  the  coefficient  of 
x\  that  is  by  8,  gives         16  x2  -24  x  =  16. 

Adding  the  square  of  the  original  coefficient  of  x  to  both  sides, 
that  is  adding  (  —  3)2,  or  9,  to  both  sides,  gives 

16z2-24z+9=25. 

The  first  member  is  now  a  perfect  square,  being  equal  to  (4  x  —  3)2. 
Therefore,  extracting  square  roots,  we  obtain 

4  x  -3  =5  and  4  x  -3  =-5. 
Solving  the  last  two  equations  gives  x=2  and  x  =  —  J.     Ans. 

EXERCISES 

Solve  each  of  the  following  quadratics  by  any  method. 

1.  9z2+6z  =  35.  4.    16z2-7o;-123  =  0. 

2.  4z2-12z  =  27.  6.    1 

3.  4z2-z-3  =  0.  6.   2 


X,  §  56]  QUADRATIC   EQUATIONS  87 


7.  2z2+6z  =  f  17.   3z2+z-200 
[HINT.    First  multiply  both  ^      r 

numbers  by  2.]  18.    x+-  =  -- 

8.  3z2-2z  =  5. 

9.  3z2+7z-110  =  0.  19.   —  --  ^  = 

r»      o       <-i  f\  3C       O  X 

10.  5z2-7z=-2. 

11.  l-3z  =  2z2.  20. 

12.  4*2-3*-2  =  0. 

13.  2a:2+3x  =  27.  21« 

14.  3x2-7x+2  =  0  [HINT.    Proceed  as  in  §  40.] 

15.  4  x2-  17  z  =-4.  22-    Vx+1  —  Va;-2  =  A/2  a;-  5. 

16.  8z  =  z2-180.  23.    Vx-l+VlQ-x=3. 

56.   Solution  by  Formula.     Every  quadratic  is  an  equation 
of  the  type  form       ax2+bx+c  =  0> 

where  a,  b,  and  c  are  given  numbers.     We  may  solve  this 
equation  as  it  stands  by  the  process  of  §  54.     Thus, 

ax?-\-bx=  —c. 
Dividing  through  by  a, 

*M*--4 

a  a 

Adding  6/(2  a)2  to  both  sides  (§  54)  gives 

x*  \bx  I  fb  b*    c 


a        \2aJ      \2aJ     a    4  a2    a        4  o2 
Extracting  the  square  root  of  both  members, 

x+  —  ==«=    /fr2-4ac^  ±Vb2-4ac 

2  a         "V     4  a2  2  a 

Transposing  the  term  6/(2  a),  we  thus  have  the  following 
formulas  for  the  two  roots : 

x.-»+vy=«a  and  x=-b-v^=47c. 

2a  2  o 


88  SECOND   COURSE    IN   ALGEBRA  [X,  §  56 

NOTE.  Observe  that  these  formulas  express  the  values  of  x 
in  terms  of  the  known  letters  a,  6,  and  c,  as  should  be  the  case.  Thus, 
in  any  example,  we  have  only  to  put  the  given  values  of  a,  6,  and  c 
into  the  formulas  in  order  to  have  at  once  the  desired  values  of  the 
two  roots. 

EXAMPLE.     Solve  by  formula  2  x2  —  3  x  =  2. 

SOLUTION.  This  equation  may  be  written  2z2—  3#—  2=0. 
Hence  the  values  of  a,  b,  and  c  in  this  case  are  as  follows:  a  =2, 
b  =  —  3,  c  =  —  2.  Placing  these  values  in  the  formulas  gives  as  the 
roots  _ 

-3)2  -4(2)  (  -2) 


2-2 

and  a,-(-3)-V(-3)«-4(2)(-2). 

2  •  2 


Simplifying,  *  = 
and  ,- 

The  two  roots  are,  therefore,  z—  2  and  x=  —  ^.  Ans.  (Com- 
pare solution  in  §  55.) 

EXERCISES 

1.  Solve  by  the  formula  Exs.  13-17,  p.  87. 

2.  Solve  by  the  formula  the  equation  3  x2—  6  x+2  =  0. 

[HINT.  The  roots  are  found  to  be  x  =  1  +^  V3  and  x  =  1  -|  Vs. 
This  quadratic  thus  has  roots  which  necessarily  contain  radicals. 
From  the  table  of  square  roots  at  the  end  of  the  book,  we  find 
V3  =  1.73205.  Hence,  the  roots,  correct  to  five  decimal  places,  are 
1+  1.73205  and  1_1.73205>  which  reduce  respectively  to  1.57735 

3  3 

and  0.42265.     Ans.] 

3.  Solve  by  the  formula  the  equation  x2  —  5  z+3  =  0,  and 
use  the  tables  if  necessary  to  determine  the  roots  decimally. 

4.  Solve  by  the  formula  the  equation  4  x2—  3  x—  2  =  0,  ex- 
pressing the  roots  decimally  correct  to  five  places. 


X,  §  56]  QUADRATIC   EQUATIONS  89 

APPLIED   PROBLEMS 

[The  method  of  solving  quadratics  by  formula  (§  56)  is  usually 
the  most  direct.] 

1.  The  square  of  a  certain  number  is  4  less   than  five 
times  the  number.     Find  the  number. 

[HINT.     Remember  that  there  should  be  two  solutions.] 

2.  Divide  20  into  two  parts  whose  product  is  96. 

3.  One  side  of  a  right  triangle  is  2  inches  longer  than  the 
other.     If  the  hypotenuse  is  10  inches  long,  how  long  are  the 
sides? 

[HINT.  Letting  x  represent  the  shorter  side,  the  equation  here 
becomes  x2  +  (x+2)2  =  100,  and  in  solving  this  we  find  that  one  of 
the  solutions  is  negative.  But  a  negative  solution  can  have  no  mean- 
ing in  such  an  example  as  this,  so  we  keep  only  the  positive  solution. 
This  frequently  happens  in  applied  problems  involving  quadratics, 
so  the  pupil  must  always  be  on  his  guard  to  keep  only  such  solu- 
tions as  can  actually  fit  a  given  problem.] 

4.  A  gardener  spades  a  bed  30  feet  long  by  20  feet  wide. 
He  then  decides  to  double  its  size  by  adding  a  border  of  uni- 
form width    throughout.     How    wide    must   ,_. 

the  border  be  made? 

5.  In  Example  4  suppose  that  instead  of 
doubling  the  area,  the  gardener  wishes  merely 
to  add  200  square  feet  to  it.     Show  that  the 

strip  added  around  the  outside  must  then  be  made  a  little 
over  1.86  feet  wide. 

6.  A  circular  swimming  pool  is  surrounded  by  a  walk  4 
feet  wide.     If  the  area  of  the  walk  is  one  fourth  that  of  the 
pool,  find  (approximately)  the  radius  of  the  pool.     (Take 


90 


SECOND   COURSE    IN   ALGEBRA 


[X,  §  56 


FIG.  14. 

is  marked  P 


A  G 


7.    If  a  train  had  its  speed  diminished  by  10  miles  an  hour, 
it  would  take  it  1  hour  longer  to  travel  200  miles.     What  is 
p     x      B  the  speed  ? 

-c  —  ^     .          L 

8.   A  circular  curbing  touches  the  line  of 
\     the  street  curbing  on  each  of  two  streets 
/     that  meet  at  right  angles.     In  Fig.  14,  the 
middle   point  of  the   circular   curbing   is 
marked   M.     The    point    at    which    the 
straight  curbings  would  meet  if  extended 
If  PM  =  5  ft.,  find  the  radius  (x  in  Fig.  14)  of 
the  circular  curbing,  f 

9.  The    figure    represents  a  pattern  fre- 
quently used  in  window  designs,  consisting  of 
a    square    A  BCD    with   a   semicircle    EFG 
mounted  upon  it,  the  diameter  GE  of  the 
semicircle  being  slightly  less  than  one  of  the 
sides  of  the  square.     If  the  shoulders  AG  and 
DE  are  each  1  foot  long,  how  long  must  each 

side  of  the  square  be  made  in  order  that  the  total  lighting 
surface  shall  be  88  square  feet? 

10.  A  soap  bubble  of  radius  r  is  blown  out  until  the  area 
of  its  outer  surface  becomes  double  its  original  value.     Show 
that  the  radius  has  thus  been  increased  by  an  amount  h  given 
by  the  formula  h  =  r(V2-l).          [HINT.    See  Ex.  14  (/),  p.  6.] 

57.  Graphical  Solution  of  Quadratics.  Consider  the 
quadratic  x2  —  3  x  —  4  =  0.  Let  us  represent  the  left  member 
by  the  letter  y  ;  that  is,  let  us  place 


FIG.  15. 


Now,  if  we  give  to  x  any  value,  this  equation  determines  a 

t  This  problem  suggests  a  practical  plan  for  finding  the  radius 
of  circular  curbings  when  the  center  0  of  the  circle  cannot  be  reached. 


X,  §  57] 


QUADRATIC    EQUATIONS 


91 


corresponding  value  for  y.  For  example,  if  x  =  Q,  then 
?/  =  02-3xO-4=-4.  Again,  if  x=l,  then  i/=l2-3xl-4 
=  —6.  The  table  below  shows  a  number  of  x-values  with 
their  corresponding  ^/-values,  determined  in  this  way. 


When  x  = 

0 

1 

2 

3 

4 

5 

.   6 

-1 

-2 

-3 

then  y  = 

4 

-6 

-6 

-4 

0 

6 

14 

0 

6 

14 

The  graph  is  now  obtained  by  draw- 
ing an  z-axis  and  a  2/-axis,  as  in  §  28, 
then  plotting  each  of  the  points  x,  y 
which  the  table  contains,  and  finally 
drawing  the  smooth  curve  passing 
through  all  such  points.  The  form  of 
the  graph  thus  obtained  is  indicated  in 
the  adjoining  figure.  Observe  that  this 
graph  is  not  a  straight  line  and  is  there- 
fore different  in  character  from  the  graph 
of  a  linear  equation.  (See  §  29.)  And 
it  is  especially  important  to  notice  that 
it  cuts  the  x-axis  in  two  points  whose 
^-values  are  —1  and  4,  respectively. 
These  two  values  of  x  determined 
in  this  purely  graphical  way  are  the 
two  solutions  of  the  given  quadratic,  x2  —  3  x  —  4  =0,  for  they 
are  those  values  of  x  that  make  y  —  0,  that  is  that  make 
z2-3z-4  =  0. 

The  graphical  study  which  we  have  just  made  of  the 
special  quadratic  x2  —  3  x  —  4  =  0  leads  at  once  to  the  following 
more  general  statements. 

Every  quadratic  has  a  graph  which -is  obtained  by  first  placing 
y  equal  to  the  left  member  of  the  equation  (it  being  understood 
that  the  right  member  is  0) ,  then  letting  x  take  a  series  of  values 


FIG.  16. 


92  SECOND   COURSE    IN  ALGEBRA  [X,  §  57 

and  determining  their  corresponding  y-values,  plotting  the 
points,  x,  y,  thus  obtained  and  finally  drawing  the  smooth  curve 
through  them. 

The  x-values  of  the  two  points  where  the  graph  cuts  the  x-axis 
will  be  the  roots  of  the  given  quadratic. 

EXERCISES 

Draw  the  graphs  of  each  of  the  following  quadratics,  and 
note  where  each  cuts  the  z-axis.  In  this  way  determine 
graphically  the  solutions,  and  check  the  correctness  of  your 
answer  by  actually  solving  by  one  of  the  methods  explained 
in  54-56. 


1.   z2-z-2  =  0.     2.   x2-7x+l2  =  Q.     3.   xz+ 

4.  z2-5z=-6. 

[HiNT.     Remember  to  write  first  as  x2  —5  x  +6  =0.] 

5.  x*+3x=W.  6.    2x2+3x  =  9. 

58.  Quadratics  Having  Imaginary  Solutions.  Consider 
the  quadratic  x2=  —  1.  This  is  a  pure  quadratic  (§  49)  and 
hence  can  be  solved  immediately  by  merely  taking  the 
square  root  of  each  member.  This  gives  as  the  required  solu- 
tions x=-\-V—  1  and  x=—\^—\.  But  V—  1  means  the 
number  whose  square  is  —1,  and  there  is  no  such  number 
among  all  those  (positive  or  negative)  which  we  have  thus  far  met. 
In  fact,  we  know  that  the  square  of  any  number,  whether  the 
number  be  positive  or  negative,  is  positive  [§  2(d)].  There- 
fore, in  any  such  case  as  this,  we  say  that  the  solutions  are 
imaginary,  and  we  speak  of  the  numbers  themselves  which, 
like  V  —  1,  enter  into  algebra  in  this  way,  as  imaginary 
numbers.  They  are  imaginary,  however,  only  in  the  sense 
that  they  have  not  been  encountered  before. 


X,  §  59]  QUADRATIC   EQUATIONS  93 

As  an  example  of  an  affected  quadratic  having  imaginary 
roots,  let  us  consider  the  equation  x2  —  6x+15  =  0.  When 
we  proceed  to  solve  this  by  the  method  of  completing  the 
square,  as  in  §  54,  the  work  is  as  follows. 

Transposing,  we  have 

x2-6x=  -15. 

Adding  9  to  both  sides  to  complete  the  square, 
z2-6z+9=  -6,  or  (z-3)2=  -6. 
Extracting  the  square  root  of  both  sides, 

x-3=  ±V^6. 
Therefore  the  solutions  are 

=3-  V^Q.    Ans. 


Both  of  these  solutions_are  seen  to  be  imaginary  because  they 
contain  the  expression  V  —Q. 

59.  Definitions.  A  number  like  3+v^6  or  3-V-6  is 
frequently  called  a  complex  number  in  distinction  to  such  a 
number  as  V  —  6,  which  is  called  a  pure  imaginary.  Thus,  a 
complex  number  is  a  combination  of  a  positive  or  negative 
number  with  a  pure  imaginary. 

All  numbers  considered  in  the  chapters  preceding  this 
(including  irrationals)  are  called  real  numbers  in  distinction 
from  the  imaginary  numbers  just  described.  Thus,  the  solu- 
tions of  all  quadratics  considered  in  §§  54-56  are  real  instead 
of  imaginary. 

EXERCISES 

Find  (by  solving)  whether  the  solutions  of  the  following 
quadratics  are  real  or  imaginary. 

1.  z2+9  =  0.  3.  2z2+2z+3  =  0.    6. 

2.  z2-6z+10  =  0.    4.  3z2+2z  =  4.          6. 


94 


SECOND    COURSE    IN   ALGEBRA 


[X,  §60 


60.  Determining  Graphically  Whether  Solutions  are  Real 
or  Imaginary.  It  was  shown  in  §  58  that  the  solutions  of 
the  quadratic  x2  —  6  x=  — 15  are  imaginary.  Let  us  now  see 
what  corresponds  to  this  fact  in  the  graph. 

The  table  below  shows  several  values  of  x  and  their  corre- 
sponding ^/-values,  as  determined  from  the  given  equation 


When  x  = 

-1 

0 

1 

w_ 

2 

3 

4 

5 

6 

then  y  = 

22 

15 

7 

6 

7 

10 

15 

FIG.  17. 


Plotting  the  various  points  (x,  y)  thus 
obtained  and  drawing  the  curve  through 
them  gives  the  graph  indicated  in  the 
accompanying  figure.  This  graph  is  es- 
sentially different  from  those  met  with 
in  §  57  in  one  particular,  namely  it  does 
not  cut  the  x-axis. 

A  similar  result  holds  for  the  graph  of 
every  quadratic  whose  solutions  are 
imaginary.  Therefore,  in  order  to  tell 
whether  the  solutions  of  any  given  quad- 
ratic are  real  or  imaginary,  we  need  only 
draw  its  graph  and  note  whether  or  not 
it  cuts  the  z-axis.  If  it  does,  the  solu- 
tions are  real ;  if  it  does  not,  the  solutions 
are  imaginary. 


EXERCISES 

Find  by  drawing  the  graph  whether  the  roots  of  each  of  the 
following  quadratics  are  real  or  imaginary. 

1.  z2+2z+3  =  0.      3.  z2-2z+3  =  0.       5.  6z2+5z+l=0. 

2.  z2+2z-3  =  0.      4.  3z2+4z+l  =  0.    6.  2z2-3z+4  =  0. 


LAGRANGE 

(Joseph  Louis  Lagrange,  1736-1813) 

Famous  for  his  discoveries  in  all  branches  of  mathematics  and  regarded  as 
the  greatest  mathematician  of  the  18th  century.  In  algebra  he  gave  much 
attention  to  the  study  of  equations  and  determinants,  extending  and  unify- 
ing the  work  of  previous  mathematicians  in  these  fields. 


PART   II.     ADVANCED  TOPICS 

CHAPTER  XI 
LITERAL   EQUATIONS   AND   FORMULAS 

61.  Literal  Equations.  Equations  in  which  some,  or  all, 
of  the  known  numbers  are  represented  by  letters  are  called 
literal  equations.  The  known  letters  are  generally  repre- 
sented by  the  first  letters  of  the  alphabet,  as  a,  b,  c,  etc. 
Literal  equations  are  solved  by  the  same  processes  as  numeri- 
cal equations,. 

EXAMPLE.     Solve  the  following  literal  equation  for  x  : 

ax  =  bx+7  c. 
SOLUTION.     Transposing, 

ax  —  bx=7  c. 
Combining  like  terms, 

(a-6)z=7c. 
Dividing  by  (a—  6), 

*=-?X     Ans. 
a—b 

CHECK.     Substituting  the  answer  for  x  in  the  given  equation, 


Multiplying  by  (a  —  b), 

7  ac  =7  be  +7  c(a  -b)  =  7  be  +7  ac  -7  be. 
Transposing, 

7  ac  —  7  ac=7  be  -7  be. 

Simplifying,  0=0,  which  is  a  correct  result. 

95 


96  SECOND   COURSE    IN   ALGEBRA          [XI,  §  61 

It  is  to  be  carefully  observed  that  a  literal  equation  is  said 
to  be  solved  for  the  unknown  letter,  as  x,  only  when  that 
letter  has  been  expressed  in  terms  of  the  other  (known) 
letters.  Thus,  in  the  example  above,  we  obtained  x  in  terms 
of  a,  6,  and  c.  This  when  once  done,  is  what  we  mean  by  the 
solution. 

NOTE.  If  a  literal  equation  is  satisfied  no  matter  what  values 
be  given  to  the  letters  appearing  in  it,  it  is  called  an  identity.  Thus 
x2  —  a2  =  (x  —  a)  (x  +a)  is  an  identity.  This  fact  is  often  expressed  by 
means  of  the  symbol  =.  Thus,  x2—  a?=(x—  o)(x+o).  Likewise, 
(z-a)2  =  z2-2  ax+a2,  etc. 


EXERCISES 

Solve  for  x  in  the  following,  checking  your  answer  in  the 
first  five. 

1.  x-a  =  b.  9   *+b  =  *+a. 

2.  ax-l  =  b.  a 

3.  ax+bx  =  c.  -       a  ,   b  _ 


4.  3z+6  =  z-3&.  cx 

5.  4(3&-z)  =  3(26+z). 

6.  (x-a)(x-b)=x(x+c). 

7.  -J--  a.  12. 


1+x  x-3 

a,  6_o  13.  Divide  a  into  two  parts 

#    #  whose  quotient  is  m. 

14.  If  A  can  do  a  piece  of  work  in  a  days,  and  B  can  do  it 
in  b  days,  how  long  will  it  take  them  working  together?  (See 
Exs.  23,  24,  p.  40.) 

[HINT.  These  are  simultaneous  equations,  to  be  solved  for  the 
two  unknowns  x  and  y  in  terms  of  a  and  b.] 


XI,  §  61]     LITERAL   EQUATIONS   AND   FORMULAS         97 


16. 


I  ax-by  =  2, 


»•{ 


3  ax-f2  by  =  ab, 
ax  —  by  =  ab. 


18. 


x    y    a 

1-1=1. 

x    y    b 

[HINT.     Solve  first  for  l/x  and  l/y.     See  Ex.  16,  p.  53.] 

a_b _  _., 
x    y~     lj 


a 


19. 

x    y 
20.  axz-c=l. 


[HINT.     See  §  51.     Ans.  x  =  * >/— •] 
21.  az2-fa3  =  5a3-3az2. 


22.  !±+±  =  i. 
x     a    x 

23. 


24.  (x+a)(x+b)+4(x+a)=Q. 
[HINT.     Solve  by  factoring.] 

25    /£2 ax  =  2  a2 

[HINT.     See  §  54,  p.  85.     Ans.  x  =2  a,  or  x  =  -a.] 

29.  x  =  4  ax — 2  a  . 

30.  Vx  —  a+Vb  — x  =  Vb  —  a. 

H 


98  SECOND   COURSE    IN   ALGEBRA          [XI,  §  62 

62.  Formulas.  If  a  person  travels  for  10  hours  at  the 
rate  of  15  miles  an  hour,  the  distance  he  travels  is  15  XlO  = 
150  miles.  Stated  in  general  (algebraic)  language,  we  can 
say  in  the  same  way  that  if  a  person  travels  for  t  hours  at  the 
rate  of  r  miles  an  hour,  the  distance  s  he  travels  is 

s  =  rt. 

This  is  a  literal  equation  expressing  the  value  of  s  in  terms 
of  r  and  t.  If  we  wish,  we  can  solve  it  for  t,  giving  t  =  s/r,  and 
what  we  now  have  is  t  expressed  in  terms  of  s  and  r.  Or,  we 
can  solve  the  original  equation  for  r,  giving  r  =  s/t,  and  this 
expresses  r  in  terms  of  s  and  t. 

These  examples  illustrate  the  important  fact  -that  in 
nearly  all  branches  of  knowledge,  especially  in  engineering, 
geometry,  physics,  and  the  like,  there  are  general  laws  which 
are  expressed  by  means  of  mathematical  formulas.  Such 
formulas  are  merely  literal  equations  in  which  two  or  more 
letters  appear,  and  it  is  often  desirable  to  solve  them  for  some 
one  letter  in  order  to  express  its  value  in  terms  of  the  others. 

EXERCISES 

1.  The  area  A  of  a  rectangle  whose  dimensions  (length 
and  breadth)  are  a  and  b  is  given  by  the  formula  A=ab. 
Solve  this  for  a ;    also  for  b.     In  each  case  state  in  terms  of 
what  letters  your  answer  is  written. 

2.  The  formula  for  the  area  A  of  a  triangle  whose  height 
(altitude)  is  h  and  whose  base  is  a  is  A  =  %  ah.     Solve  for  a ; 

also  for  h. 

3.   Solve  for  b  in  the  formula 


(Formula  for  the  area  A  of  a  trapezoid  whose 
FIG.  is.  bases  are  B  and  b  and  whose  altitude  is  h.) 


XI,  §62]      LITERAL   EQUATIONS   AND   FORMULAS        99 

4.  Solve  for  r  in  the  formula  A=irr2.     (Formula  for  the 
area  of  a  circle  whose  radius  is  r.) 

5.  The  interest  /  which  a  principal  of  p  dollars  will  yield 
in  t  years  at  r  %  is  determined  by  the  formula 


100 

Solve  this  for  r  and  use  your  result  to  answer  the  following 
question  :  What  rate  of  interest  is  necessary  in  order  that  $50 
may  yield  $6  interest  in  2  years'  time  ? 
[HiNT.     Solving  for  r  gives  at  once 
r  =  1007 
pt   ' 

Now  see  what  the  right  member  of  this  equation  becomes  when 
7=6,  p=50,  and  t=2.] 

6.  Using  tne  interest  formula  of  Ex.  5,  solve  it  for  t  and 
use  your  result  to  answer  the  following  question  :     How  long 
will  it  take  $600  to  yield  $63  interest  if  invested  at  6%? 

7.  The  velocity  of  sound  v,  in  feet  per  second,  is  given 
by  the  formula  v=  1090+1.  140-32),  where  t  is  the  tem- 
perature of  the  air  in  Fahrenheit  degrees.     Find 

(a)  The  velocity  of  sound  when  the  temperature  is  75°. 

(6)  The  temperature  when  sound  travels  1120  ft.  per  sec. 

8.  Derive  formulas  for  each  of  the  following  statements. 
(a)  The  number  N  of  turns  made  by  a  wagon  wheel  d  feet 

in  diameter  in  traveling  s  miles. 

(6)  The  number  N  of  dimes  in  m  dollars,  n  quarter  dollars, 
and  q  cents. 

9.  An  automobile  travels  for  T  hours  at  the  rate  of  v 
miles  per  hour.     By  how  much  must  this  rate  be  increased 
in  order  to  make  the  same  journey  in  t  minutes  less  time? 

10.  A  has  $a  and  B  has  $6.  Between  them  they  give  $c 
to  a  certain  charity,  after  which  the  amounts  of  money  they 
have  are  equal.  How  much  does  each  contribute? 


100 


SECOND   COURSE    IN   ALGEBRA          [XI,  §  63 


63.   Law  of  the  Lever.     If  two  weights  are  balanced  at  the 
ends  of  any  (uniform)  bar,  as  shown  in  the  figure,  we  have  an 

example  of  a  lever.  The  point  of 
support,  F,  is  called  the  fulcrum. 
If  we  let  W  and  w  be  the  values 
(in  pounds  or  ounces  or  any 
FIG  19  other  convenient  unit)  of  the  two 

weights,  while  D  and  d  stand  for 

the  distances  respectively  of  W  and  w  from  F,  then,  when- 
ever the  balance  is  perfect,  we  have  the  formula 


D         F              d 

A 

fw 

W 

w    D 

Sometimes  a  single  weight  W  is  balanced  by  a  force,  p, 
usually  called  a  power.  This  may  happen  in  several  ways,  as 
indicated  by  the  following  figures.  In  all  such  cases,  if  we 


I— D- 


(2) 


let  W  represent  the  weight,  p  the  power,  D  the  distance  from 
W  to  the  fulcrum,  there  exists  the  following  formula  when- 
ever the  balance  is  perfect : 

W=d 
P     D 


XI,  §64]    LITERAL   EQUATIONS   AND   FORMULAS        101 


This  is  called  the  general  law  of  the  lever.  By  clearing  the 
equation  of  fractions,  it  may  be  written  in  the  form 

WD=pd. 

Translated  into  words,  this  last  relation  means  that  the 
weight  times  the  weight  arm  equals  the  power  times  the  power 
arm.  It  is  in  this  form  that  the  law  is  usually  remembered  by 
engineers. 

EXERCISES   ON   THE   LEVER 

1.  If  the  fulcrum  of  a  5-foot  crowbar  is  placed  1  foot  from 
the  end,  what  weight  can  be  lifted  by  a  man  weighing  180 
pounds  ? 

[HINT.  Here  we  have  Fig.  19  with  W  =  ?,  w  (or  p)  =  180  pounds, 
D=lfoot,  d  =4  feet.] 

2.  The  figure  represents  a  simple 
form  of  pump.     If  the  pump  handle 
AF  is   16   inches  long,   while  the 
piston-arm  FC  is  3  inches  long,  what 
will  be  the  upward  pull  at  C  when 
there  is  a  9-pound  downward  push 
at  A? 

3.  A  certain  lever,  after  being  balanced,  has  r  pounds 
added  to  the  weight  W.     Determine  (in  terms  of  W,  p,  D,  d, 
and  r)  how  much  the  power,  p,  must  be  increased  to  keep  the 
balance  perfect. 

64.  Gear  Wheel  Law.  Whenever  a  gear  wheel  having  T 
teeth  revolves  at  the  rate  of  N  revolutions  per  minute  and 
turns  another  similar  wheel  having  t 
teeth  at  the  rate  of  n  revolutions  per 
minute,  there  exists  at  all  times  dur- 
ing the  motion  the  formula 

T=n. 
Fro.  22.  t      N 


FIG.  21. 


102  SECOND   COURSE    IN   ALGEBRA          [XI,  §  64 

This  is  the  gear  wheel  law.  Clearing  this  equation  of  frac- 
tions, it  becomes 

TN=tn. 

Translated  into  words,  this  last  relation  means  that  the 
number  of  teeth  in  one  wheel  multiplied  by  its  rate  of  turning  is 
equal  to  the  number  of  teeth  in  the  other  wheel  multiplied  by  its 
rate  of  turning. 

EXERCISES    ON    GEAR    WHEELS 

1.  If  in  Fig.  22  the  small  wheel  has  30  teeth  and  is  making 
96  revolutions  per  minute,  how  many  teeth  must  the  large 
wheel  have  in  order  to  revolve  16  times  per  minute? 

2.  In  order  that  the  large  wheel  revolve  three  fourths  as 
fast  as  the  small  one,  how  must  the  wheels  be  made  ? 

3.  If  the  large  wheel  in  §  64  be  made  larger  by  the  addi- 
tion of  r  teeth  to  its  run,  determine  the  amount  by  which  the 
speed  of  the  smaller  wheel  will  be  thereby  increased. 

[HINT.  Let  x  represent  the  unknown  amount  and  find  a  formula 
for  x  in  terms  of  T,  t,  N,  n,  and  r.] 

65.   Other  Useful  Formulas.     In  addition  to  the  formulas 
already   mentioned,    the   following   from   plane   and   solid 
geometry  and  from  elementary  physics  are 
often  used. 

1.   The  area  A  of  an  equilateral  triangle 
of  side  a  (Fig.  23)  is 

_V3    2  

FIG.  23.  :~^~fl' 

where  A/3  =  1.732  approximately. 

2.   The  area  A  of  a  regular  hexagon  of 
side  a  (Fig.  24)  is 

3\/3 


fl2. 


2  FIG.  24. 


XI,  §  65]     LITERAL   EQUATIONS   AND    FORMULAS       103 


3.  The  area  A  of  any  triangle  in 
terms  of  its  three  sides  a,  6,  and  c 
(Fig.  25)  is  

A  =  Vs(s-a)(s-b)(s-c), 
where  s  = 


a 
FIG.  25. 


4.  The  area  A  of  the  sector  of  a  circle  when 
the  intercepted  arc  is  a  and  the  radius  is  r 
(Fig.  26),  is 


FIG.  26. 


5.   The  length  of  the  diagonal  d  of  a 
rectangular    block    whose     dimensions    b\ 
are  a,  6,  and  c  (Fig.  27)  is 


6.  The  volume  V  of  a  circular  cylinder  of  altitude 
h  and  radius  of  base  r  (Fig.  28)  is 

F=irr2/i. 

7.  The   volume    V   of   a   circular 
cone  of  altitude  h  and  radius  of  base  r  (see 

Fig.  29)  is 


FIG.  28. 


V=±Trr*h. 

8.   The  volume  V  of  a  pyramid 
of  altitude  h  and  base  B  (Fig.  30)  is 


9.  The  volume  V  of  a  spheri- 
cal segment  (or  slice  of  a  sphere 
between  two  parallel  cutting  planes),  where 
h  is  the  altitude,  and  a  and  b  the  radii  of  the 
two  bases  (Fig.  31)  is 


Fio.  29. 


FIG.  30. 


FIG.  31. 


104 


SECOND   COURSE    IN   ALGEBRA 


[XI,  §  65 


10.  The  surface  S  of  the  zone  (or  portion  of  the  surface  of 
a  sphere  lying  between  two  parallel  cutting  planes),  where 
h  is  the  distance  between  the  cutting  planes  and  r  the  radius 

of  the  sphere  (Fig.  31),  is 

S  =  2*117/1. 

11.  The  length  of  belt  I  required  to  go  around  two  wheels 
whose  diameters  are  D  and  d  and  whose  centers  are  at  the 
distance  a  apart  is 


FIG.  32. 

(a)  In  case  the  belt  does  not  cross  itself, 


(6)  In  case  the  belt  crosses  itself  once, 


/  = 


2 


12.   The  force  F,  measured  in  pounds,  with  which  a  body 
weighing  W  pounds  pulls  outward  (centrifugal  force)  when 
traveling  with  a  velocity  of  v  feet  per 
second  in  a  circle  of  radius  r  (Fig.  33) 
is  determined  by  the  formula 


FIG.  33. 


„ 

r 


32 


13.  The  pressure  P  exerted  by  a  letter 
press  (Fig.  34)  is  determined  by  the 
formula  D  2irrF 


where  F  is  the  value  of  the  force  applied  at  FIG.  34. 


XI,  §  65]     LITERAL   EQUATIONS  AND   FORMULAS       105 

the  wheel,  r  is  the  radius  of  the  wheel,  and  h  is  the  distance 
from  one  thread  of  the  screw  to  the  next  one. 

14.  The  weight  W  which  can  be  raised 
by  means  of  a  toothed  wheel  and  screw 
such  as  indicated  in  Fig.  35  is  deter- 
mined by  the  formula 


where  P  represents  the  pressure  applied 

at  the  handle  and  where  R,  r,  d,  and  I  are  the  dimensions 

indicated  in  the  figure. 

Other  important  formulas  from  elementary  geometry  and 
from  physics  may  be  found  in  Chapter  XVII. 

EXERCISES 

1.  Find  by  Formula  3  of  §  65  the  area  of  the  triangle  whose 
three  sides  are  respectively  5  inches,  5  inches,  and  8  inches 
long. 

2.  Show  that  in  the  case  of  an  equilateral  triangle  of  side 
a  the  expression  for  A  in  Formula  3,  §  65,  reduces,  as  it  should, 
to  that  for  A  in  Formula  1,  §  65. 

3.  Show  that  in  case  the  two  wheels  in  Fig.  32  have  the 
same  diameter  D,  the  formula  for  the  length  of  belt  re- 
duces in  case  (a)  to  the  simple  form  l=TrD+2  a,  and  in  case 
(6)  to  Z  =  2VD2+a2+7rD. 

4.  How  much  leather  (surface  measure)  will  it  take  for  a 
belt  6  inches  wide  to  connect  two  pulleys  whose  diameters 
are  5  feet  and  1  foot,  respectively,  the  distance  between 
centers  being  10  feet  ? 

[HINT.     Viol  =  10.2,  approximately.] 


106  SECOND   COURSE    IN   ALGEBRA          [XI,  §  65 

5.  A  pail  of  water  weighing  5  pounds  is  swung  round  at 
arm's  length  at  a  speed  of  10  feet  per  second.     If  the  length 
of  the  arm  is  2  feet,  find  (a)  the  pull  at  the  shoulder  when 
the  pail  is  at  the  uppermost  point  of  its  course,  (b)  when  at 
the  lowest  point  of  its  course.     Also,  find  the  least  velocity 
which  the  pail  can  have  without  the  water  dropping  out  at 
the  top  point  of  the  course. 

6.  What  pressure  is  exerted  by  a  letter  press  in  case  the 
force  applied  at  the  wheel  is  10  pounds,  the  diameter  of 
the  wheel  is  1^  feet,  and  the  threads  of  the  screw  are  %  inch 
apart  ? 

7.  In  the  device  shown  in  Fig.  35,  show  that  if  the  distance 
d  between  two  adjacent  threads  be  halved  and  the  number 
of  teeth  on  the  wheel  be  correspondingly  doubled  to  fit  the 
new  gear,  other  parts  remaining  the  same,  the  weight  W  that 
can  be  raised  with  a  given  pressure  P  will  be  doubled. 

8.  The  volume  V  of  the  frustum  of  a  cone  or  pyramid 
made  by  a  plane  parallel  to  the  base  is  given  by  the  formula 


where  B  and  b  denote  the  areas  of  the  lower  and  upper  bases, 
and  h  denotes  the  altitude. 


FIG.  36. 


Determine  the  volume  of  the  frustum  of  a  circular  cone  the 
radii  of  whose  bases  are  4  inches  and  3  inches,  the  altitude 
being  5  inches. 


XI,  §  65]      LITERAL   EQUATIONS  AND   FORMULAS       107 

Show  that  in  case  the  upper  and  lower  bases  are  equilateral 
triangles  whose  sides  are  a  and  b  respectively,  the  formula 
becomes 


[HINT.     See  Formula  1,  §  65.] 

9.  Show  by  means  of  Formula  3,  p.  103,  that  if  the 
three  sides  of  any  triangle  be  extended  to  twice  their  original 
length,  the  area  will  become  quadrupled. 

10.  If,  in  the  first  of  the  two  cases  represented  in  Fig.  32, 
the  diameter  of  each  wheel  be  increased  by  the  same  amount, 
say  a,  show  that  the  length  of  belt  will  thereby  become  in- 
creased by  the  amount  ira. 

11.  If,  in  the  second  of  the  cases  represented  in  Fig.  32, 
the  diameter  of  the  large  wheel  be  increased  by  any  given 
amount  and  at  the  same  time  the  diameter  of  the  small 
wheel  be  diminished  by  the  same  amount,  show  that  pre- 
cisely the  same  length  of  belt  as  before  will  fit  over  them 
tightly. 

12.  A  body  is  moving  along  a  circle  with  a  velocity  of 
3'  feet  per  second.     Show  that  in  order  for  the  centrifugal 
force  (12,  p.  104)  which  it  is  exerting  to  be  doubled,  the 
velocity  must  be  increased  by  about  1J  feet  per  second. 

By  how  much  would  the  velocity  have  to  be  diminished 
in  order  that  the  centrifugal  force  become  halved  ? 

13.  By  means  of  Formulas  9  and  10  (§  65)  obtain  the 
formulas  for  the  area  and  volume  of  a  whole  sphere.     Com- 
pare with  Ex.  14,  (e),  (/),  p.  6. 


CHAPTER  XII 
GENERAL  PROPERTIES  OF  QUADRATIC  EQUATIONS 

66.  The  Classification  of  Numbers.    A  real  number  is 

one  whose  expression  does  not  require  the  square  root  of  a 
negative  quantity,  while  an  imaginary  number  is  one  whose 
expression  does  require  such  a  square  root.  (See  §  59,  p.  93.) 

Thus  1,  3,  -7,  £,  f,  -f  ,   \/2,  1+  \/3  are  real  numbers,  while 
-S,  V-  i,  2  +  A/-3,  are  imaginary  numbers. 

In  case  a  real  number  can  be  expressed  in  the  particular 
form  ^  where  p  and  q  are  integers  (positive  or  negative)  it  is 

q 

called  a  rational  number.  The  number  zero  is  also  included 
among  the  rational  numbers.  (See  §  42,  p.  56.) 

Thus  l,  f  ,  —  f  ,  5,  73,  —10,  —  ^J,  are  rational  numbers. 

In  case  a  real  number  cannot  be  expressed  in  the  particu- 
lar form  just  mentioned,  it  is  called  an  irrational  number. 


Thus  V2,  V3,  Vf  ,  ^2,  -vXJ",  V^  1  +  A/6,  are  irrational  num- 
bers. 

Imaginary  numbers  are  either  pure  imaginaries,  such  as 
V^3,  or  complex  numbers  (§  59,  p.  93),  such  as  1  +  V  —  3. 

108 


XII,  §  67]    GENERAL  PROPERTIES  OF  QUADRATICS    109 

These  divisions  and  subdivisions  may  be  summarized  into 
a  table  as  below : 

n     ,  f  Rational 

Real{  T      ..      , 

»r      »  f   .,  Irrational 

Numbers  of  Algebra  T 

.        .         j  Pure  Imaginanes 

Imaginary  {  „       1      ,yr      , 

\  Complex  Numbers 

NOTE.  The  rational  numbers  are  themselves  subdivided  into 
three  sub-classes :  the  single  number  zero,  the  integers  (positive  or 
negative),  and  the  rational  fractions.  The  latter  are  those  rational 
numbers  which,  like  f ,  cannot  be  expressed  as  integers. 

All  the  numbers  used  in  arithmetic  are  positive  real  numbers. 
The  negative  real  numbers  together  with  the  imaginary  numbers 
owe  their  existence  to  algebra. 

67.  Determining  the  Character  of  the  Roots  of  a  Quad- 
ratic. It  is  often  desirable  to  determine  the  character  of 
the  roots  of  a  given  quadratic,  that  is,  whether  the  roots  are 
real  or  imaginary ;  and  if  real,  whether  they  are  rational  or 
irrational,  etc. 

Thus  the  roots  of  2  z2-7  x+l  =  0  are  (by  §56,  p.  87),  7=t=V^. 

4 

Since  41  is  positive,  these  roots  are  real  numbers  (§  66). 

Since  41  is  not  a  perfect  square,  the  roots  are  irrational  (§  66). 

Since  VH  is  added  to  7  in  the  one  root  and  subtracted  from  7  in 
the  other  root,  the  roots  are  unequal. 

Thus  the  character  of  the  roots  in  this  case  is  described  by  saying 
that  they  are  real,  irrational,  and  unequal. 

There  is,  however,  a  much  shorter  method  than  the  one 
illustrated  above  for  determining  the  character  of  the  roots 
of  a  given  quadratic.  Thus  we  know  (§  56)  that  the  two 
roots  of  any  quadratic,  namely,  any  equation  of  the  form 

ax2-\-bx-\-c  =  Q, 


_       and  _ 

2a  2a 


110  SECOND   COURSE    IN   ALGEBRA         [XII,  §  67 

An  examination  of  the  form  of  these  expressions  gives 
at  once  the  following  rule. 

RULE.  For  any  given  quadratic,  ax*+bx+c  =  Q,  in  which 
the  coefficients  a,  b,  c  are  real  numbers,  the  two  roots  will  be 

(1)  Real  and  unequal  if  b2  —  4  ac  is  positive. 

(2)  Real  and  equal  if  b2  —  4  ac  =  0.      (Both    roots    then 
reduce  to  —  6/2  a.) 

(3)  Imaginary  ifb2  —  4acis  negative. 

Moreover,  if  the  coefficients  a,  b}  c  are  rational  numbers,  the 
two  roots  will  be 

(4)  Rational,  if  62  —  4  ac  is  a  perfect  square;  irrational  if 
b2—4ac  is  not  a  perfect  square. 

Because  of  the  manner  in  which  the  character  of  the  roots 
thus  comes  to  depend  upon  the  value  of  b2  —  4  ac,  this  expres- 
sion is  called  the  discriminant  of  the  given  quadratic. 

EXAMPLE  1.  Determine  the  character  of  the  roots  of 
2z2-3z-9  =  0. 

SOLUTION.  Here  a  =2,  b  =—3,  c=  —  9.  Hence  the  value  of 
the  discriminant,  or  62-4ac,  is  (  -3)2-4(2)(  -9)  =9+72=81  =92. 

Hence,  by  (1)  and  (4)  of  the  rule,  the  roots  are  real,  unequal, 
and  rational. 

EXAMPLE  2.     Determine  the  character  of  the  roots  of 


SOLUTION.   Here  a  =3,  b=2,  c  =  l.    Hence  62—  4  ac=4  —  12=  -8. 
Hence,  by  (3)  of  the  rule,  the  two  roots  must  be  imaginary. 

EXAMPLE  3.     Determine  the  character  of  the  roots  of 
4z2-20z+25  =  0. 

SOLUTION,   a  =4,  6=  -20,c=25.    Hence  62-  4  ac=  400  -400=0. 
Therefore,  by  (2)  of  the  rule,  the  roots  are  real  and  equal. 
The  common  value  which  the  two  roots  have  may  be  found  if 
desired  by  actually  solving  the  equation.     It  turns  out  to  be  f  . 


XII,  §68]   GENERAL  PROPERTIES  OF  QUADRATICS    111 


EXERCISES 

Determine  (without  solving)  the  character  of  the  roots  of 
each  of  the  following  equations. 

1.   2z2-3z+l=0.  7. 


x*+x=-l. 


8. 

9. 
10. 
11. 
12. 


*68.   Character  of  Roots  Considered  Geometrically.    We 

have  seen  in  §  57  that  whenever  a  quadratic  has  its  two  roots 
real  and  unequal,  its  graph  will  cut  the  #-axis  in  two  distinct 
points.  On  the  other  hand,  if  the  roots  are  imaginary,  the 
graph  of  the  equation  will  not  cut  the  a>axis  at  all  (§  60). 
Suppose  now  that  we  have  a  quadratic 
whose  two  roots  are  equal  to  each  other, 
for  example 

(1)  4z2-12z+9  =  0. 

Here  the  discriminant  is 

(-12)2- 4(4)  (9)  =  144 -144  =  0, 
so  that  the  roots  must  be  equal  (§  67). 
If  we  now  proceed  to  draw  the  graph 
of   (1)   in   the   usual  way   by   placing 
2/  =  4z2— 12  z+9,  then  forming  a  table 
of  x,  y  values,  etc.,  it  appears  that  the  . 
graph  corresponding  to  (1)  just  touches 
the  x-axis  instead  of  cutting  through  it. 
This,  in  fact,  is  what  we  should  expect, 
since  the  equality  of  the  roots  virtually 
means  that  there  is  but  one  root,  and  this  can  be  possible 
only  when  the  graph  merely  touches  (is  tangent  to)  the  x-axis. 


FIG.  37. 


112 


SECOND   COURSE    IN   ALGEBRA         [XII,  §  68 


In  order  to  illustrate  in  one  single  diagram  all  the  facts 
mentioned  thus  far  about  the  graph,  it  is  instructive  to  take 
such  an  equation  as 
(2)  z2-2z+c  =  0, 

and  let  c  take  different  values,  thus  obtaining  various  quad- 
ratics. For  example,  if  we  choose  c—  —  2,  the  quadratic 
equation  becomes 


and  if  we  draw  the  graph  of  this,  we  find  that  it  cuts  through 

the  z-axis  at  two  points,  thus  in- 
dicating that  its  solutions  are  real 
and  unequal. 

Likewise,  we  find  a  similar  result 
when  we  let  c=—  1,  and  when 
c  =  0,  though  the  various  graphs  are 
themselves  different. 

But  if  we  choose  c  =  l  and  pro- 
ceed as  before,  the  graph  of  the 
corresponding  quadratic  no  longer 
cuts  through  the  o>axis,  but  merely 
touches  it,  thus  indicating  real 
and  equal  roots. 

Finally,  for  such  values  of  c  as  2, 
3,  or  4,  the  quadratics  (2)  come 
to  have  graphs  which  do  not  cut 
the  x-axis,  thus  indicating  that  they  have  imaginary  roots. 
The  effect  of  changing  c  is  thus  merely  to  slide  one  and 
the  same  curve  vertically  up  and  down  the  coordinate  paper. 
The  figure  shows  the  positions  of  the  curve  corresponding 
to  c  =  —  2,  c  =  1  ,  and  c  =  4.  The  pupil  is  advised  to  draw 
in  for  himself  the  positions  of  the  curve  for  c  =  —  1,  c  =  0,  c  =  2, 
and  c  =  3. 


FIG.  38. 


XII,  §69]    GENERAL  PROPERTIES  OF  QUADRATICS    113 

69.  The  Sum  and  Product  of  the  Roots.    We  have  seen 
that  every  quadratic  is  an  equation  of  the  form 

(1) 


and  we  have  also  seen  (§  56)  that  the  roots,  or  solutions,  of 
this  equation  are 


and 


-6-V&2-4 


oc 


2a 

It  is  now  to  be  observed  that  if  we  add  these  solutions 
together  the  radical  cancels  and  we  obtain  the  simple  result 

-2b_     b 
2  a          a 

Again,  if  we  multiply  the  two  solutions  together,  we  obtain 
a -final  result  which  is  very  simple  in  form.     Thus 

_  (-fr)2-  (Vb2-4  ac)2_62-  (62-4  ac)  _4  ac_  c 
4  a2  4  a2          ~4~a?~a' 

These  results  may  be  summarized  in  the  following  rule. 

RULE.     In  the  general  quadratic  equation  ax2-\-bx-\-c  =  Q, 

(1)  The  sum  of  the  two  solutions  is  —  b/a. 

(2)  The  product  of  the  two  solutions  is  c/a. 

EXAMPLE.     Find  the  sum  and  the  product  of  the  solutions 
of  the  equation  3x2  —  2  z+6  =  0. 
SOLUTION.     Here  a  =  3,  6=  —2,  c  =  6. 

r> 

Hence  the  sum  of  the  solutions  is ,  or  f,  and  the  product 

of  the  solutions  is  %  =  2. 
l 


114  SECOND   COURSE    IN   ALGEBRA        [XII,  §  69 

EXERCISES 

State  (by  inspection)  the  sum  and  the  product  of  the  solu- 
tions of  each  of  the  following  equations.  Check  your  answer 
in  Exs.  1,  2,  3,  4  by  actually  solving  the  equations  and  de- 
termining the  sum  and  the  product  of  the  solutions. 

1.  2z2+5z-7  =  0.  5.   6z2+7z  =  42. 

2.  3x*-7x+2  =  Q.  6. 

3.  5z2-2z  =  16.  7. 

4.  3z  =  200-z2.  8.   x2+px  =  q. 

9.  Show  that  in  the  quadratic  x2 -\-rnx +n  =  0  the  sum  of  the 
roots  is  —  m  and  the  product  of  the  roots  is  n.     This  general 
result  is  important  and  may  be  stated  in  words  as  follows : 

//  in  a  quadratic  the  coefficient  of  x2  is  1,  the  sum  of  the  solu- 
tions will  be  the  coefficient  of  x  with  its  sign  changed,  while  the 
product  of  the  solutions  will  be  the  remaining  term.  Explain 
and  illustrate  by  means  of  the  equation  z2— 10  z+12  =  0. 

10.  Apply  the  result  stated  in  Ex.  9  to  determine  the  sum 
and  the  product  of  the  solutions  of  the  following  equations: 

(a)  z2-5z+7  =  0.  (e)  x2-(a+b)x+ab  =  Q. 

(b)  z2-4z  =  10.  (/)   2z2+3z-4  =  0. 

(c)  x2  —  %x  =  2.  [HINT.     First  divide  through 

(d)  z2-V2z+V3  =  0.  by  2.] 

70.   Formation  of  Quadratics  Having  Given  Solutions. 

EXAMPLE  1.  Form  the  quadratic  -whose  solutions  are 
1  and  -i-. 

SOLUTION.     If  x  =  l,  then  x  -1=0;  if  x  =  -J-,  thenz+£  =  0. 

Hence  the  equation  (x-l)(z+£)  =0,  or  z2-^  x-±=Q,  will  be 
satisfied  when  either  z  =  lorz=  —  ^-(§  52). 

The  desired  quadratic  is  therefore 

z2  — 1-  x  — J-  =0,  or  2  x2  -x  —  1  =0.     Ans. 


XII,  §70]    GENERAL  PROPERTIES  OF  QUADRATICS    115 

Similarly,  if  the  given  solutions  are  any  numbers,  as  a 
and  6,  the  equation  having  these  as  solutions  is  obtained  by 
subtracting  a  from  x  and  6  from  x,  then  multiplying  the  two 
factors  thus  obtained  together  and  placing  the  result  equal 
to  0 ;  that  is,  the  desired  equation  is  (x  —  a)(x  —  b)  =  0» 


EXERCISES 

Form  the  quadratics  whose  roots  are 

1.  2,  3.  8. 

2.  -2,  -3.  9.  3m,  -5m. 

3.  4,  f  10.  2a-6,  2  a+b. 

4.  12,  -5.  11.  3+V2,  3-V2. 

5.  i,  -£.  12.  2-V5,  2+ V5. 

6.  V2,  V3.  13.  2±V3. 
[HINT.      Write  answer  in  the  14.  —  \  (3  ±  V§) . 

form  *2-(V2 +  V3)z  +  V6  =0.]  lg  i(_i±V2). 

7.  V8,  -\/2.  16.   m(2±2V5). 

17.  Show  that  if  in  any  quadratic,  as  az2+6z+c  =  0,  one 
root  is  double  the  other,  then  the  relation  2  b2  =  9  ac  must 
exist  among  the  coefficients,  a,  6,  and  c. 

[HINT.  Let  r  be  one  root.  Then,  by  what  the  problem  assumes, 
the  other  root  is  2  r.  Now  form  the  quadratic  having  r  and  2  r  as 
roots,  and  examine  its  coefficients.] 

18.  Show  that  if  in  any  quadratic,  as  ax2 -\-bx-\- c  =  0,  one 
root  is  three  times  the  other,  then  we  must  have  4  ac  =  62  —  9  a2. 

19.  Find  the  relation  which  must  hold  between  a,  b,  and  c 
in  order  that  the  roots  of  the  quadratic  ax2+bx+c  =  Q  may 
be  to  each  other  in  the  ratio  2:3. 

[HINT.    Use  the  Rule  of  §  69.] 


CHAPTER  XIII 
IMAGINARY   NUMBERS 

71.  Preliminary  Statement.    Just  as  V2  means  the  num- 
ber whose  square  is  2 ;  that  is,  (V2)2  =  2,  so  V— 2  means  the 
number  whose  square  is  —2;  that  is,  (V  — 2)2=  —2.     The 
latter  case  differs  essentially,  however,  from  the  former,  be- 
cause we  cannot  conceive  of  any  number,  positive  or  nega- 
tive, whose  square  gives  a  negative  result,  like  —2.     In  fact, 
this  would  seem  to  contradict  the  law  of  signs  [§  2  (d)], 
according  to  which  the  square  of  either  a  positive  or  negative 
quantity  is  always  positive.     The  explanation  is  that  V— 2 
belongs  to  an  altogether  new  class  of  numbers,  called  im- 
aginary numbers,  so  that  we  cannot  expect  to  think  of  them 
in  the  way  just  mentioned  ;  namely,  as  though  they  were  real 
numbers.     Imaginary  numbers  first  came  to  our  notice  in 
§  58,  where  we  found  that  the  very  simple  quadratic  x2  =  —  1 
has  the  two  imaginary  roots  x  =  V—  1  and  x—  —  V— 1. 

Imaginary  numbers  have  certain  definite  properties. 
They  may  be  added,  subtracted,  multiplied,  divided,  etc., 
in  ways  which  will  be  explained  in  the  present  chapter. 

72.  The  Imaginary  Unit.     Every  pure  imaginary  number 
(§  66)  may  be  expressed  as  the  product  of  a  certain  real  num- 
ber multiplied  by  V— 1.     For  this  reason,  V— 1  is  called 

116 


XIII,  §  72] 


IMAGINARY  NUMBERS 


117 


the  imaginary  unit,  and  for  convenience  is  represented  by 
the  letter  i. 


Thus    V^ 


V27(  -  1)  =  V27  V^I  = 


A  pure  imaginary  number  when  thus  written  as  a  real 
number  multiplied  by  i  is  said  to  be  expressed  in  terms  of  i. 

EXERCISES 

Write  each  of  the  following  expressions  in  terms  of  i. 

5.    V^ 


1.  V-IQ. 

2.  V-25. 


3.    V— 18. 

SOLUTION  OF  Ex.  9. 

10.  v^J". 

11.  V^T. 


12. 


13.        - 


14. 

SOLUTION  OF  Ex.  14. 


ll^  J?7  V^ 
4       '  4 


i. 


Ans. 


16.      = 


118 


SECOND    COURSE    IN   ALGEBRA       [XIII,  §  73 


73.   Addition  and  Subtraction  of  Pure  Imaginary  Num- 
bers. 

EXAMPLE.     Add  V^9  and  V-25  and  express  the  result 
in  terms  of  i. 
SOLUTION. 


EXERCISES 

Simplify  each  of  the  following  expressions,  obtaining  the 
result  in  terms  of  i. 


-81  +     -64+-100. 


-16a2+ ~-100a2-  V-81a2. 


9.  V  -  25  zy  +  v  -  225  x*y2+  V  -  625  x2y2- 

10.  V-32m2-hV-20m2-V-27m2. 

11.  V-24  /i2/ 

12.  V 


74.   Simplification  of  Complex  Numbers. 
EXAMPLE.     Simplify  2+V-9. 

SOLUTION.     24-^^9=2+3^.     ^Ins. 

In  this  exercise  we  have  a  real  number,  2,  to  which  is  added  the 
pure  imaginary  number,  V  —  9,  thus  giving  a  complex  number  (  §  59). 

EXERCISES 

Simplify  each  of  the  following  complex  numbers. 
1. 


2.   6-2V 
3- 


XIII,  §75]  IMAGINARY   NUMBERS  119 

7.  Show  that  the  quadratic  equation  x2  —  4z+13  =  0  has 
as  its  solutions  x  =  2  ±  3  i. 

[HINT.      Solve  as  in  §  56.] 

8.  Find  the  solutions  of  the  equation  4(2  x  —  5)  =x2. 

75.   Multiplication  of  Imaginary  Numbers. 
EXAMPLE  1.     Multiply  V^3  by  V^4. 
SOLUTION.     V^3  •  V^4  =  V$i  •  V±i  =  Vl2-  i*=  Vl2(V-l)2 

=  -Vl2=-2V3.     Ans. 

Note  that  the  process  of  multiplication  consists  in  first 
expressing  each  number  in  terms  of  the  unit  i,  then  making 
use  of  the  fact  that  (since  i  =  V  —  1)  we  may  write  —  1  for  i2. 


EXAMPLE  2.     Multiply  2+^3  by  4-~^3. 
SOLUTION.     (2  +  V^3)  (4  -  V~^3)  =  (2  +  Vs  i)  (4  -  VJTt). 


8+4  V3i 


8+2  V3i-3(-l)=  11+2  V3i.     Ans. 


EXERCISES 

Find  each  of  the  following  indicated  products. 

9. 
10. 
11. 
12. 
13. 
14. 

15.    a 
16. 


120  SECOND   COURSE    IN   ALGEBRA       [XIII,  §  76 

76.   Division  of  Imaginary  Numbers. 
EXAMPLE  1.     Find  the  value  of  V^27-r-  V^3. 


so—      --    ---  -* 


EXAMPLE  2.     Find  the  value  of  6  -^V^3. 

SOLUTION.     _^=-^_  =_^_  =^_=^M=  _2Vs  i. 
^       V         V2     -V 


EXAMPLE  3.     Find  the  value  of  3  -5-  (1  +  V^~2)  . 
SOLUTION.  —2  ---  L—  -  3(1  -V2i) 


.    ^ 


l-2i2  1+2 

Note  that  in  Example  2  we  rationalized  the  denominator 

of  -  %—  by  multiplying  both  numerator  and  denominator 

-Va 

by  the  value  A/3.     Likewise,  in  Example  3  we  rationalized 

Q 

the  denominator  of  -     —  —  -  by  multiplying  both  numerator 


and  denominator  by  1  —  V2t.  In  general,  any  fraction  hav- 
ing a  denominator  of  the  form  a-\-  v  —  b  may  be  rationalized 
by  multiplying  both  numerator  and  denominator  by  a—  V  —  6, 
which  is  called  the  conjugate  imaginary  of  a+ V— 6. 

EXERCISES 

Find  the  following  indicated  quotients,  rationalizing  the 
denominator  in  each. 

4.  2-i-V^6. 

5.  V%  -r-  V^2. 


IMAGINARY  NUMBERS 


121 


11. 


*77.  Geometric  Representation  of  Complex  Numbers.  All  real 
numbers  (positive  or  negative)  can  be  represented  as  points  on  a  line, 
as  explained  in  §  1.  Similarly,  all  complex  numbers  may  be  repre- 
sented as  points  in  a  plane,  and  it  is  convenient  for  many  purposes 
to  regard  them  in  this  way.  Thus, 
the  complex  number  5  +4  i  may  be 
looked  upon  as  lying  at  the  point 
(5, 4),  that  is  at  the  point  whose 
abscissa  is  5  and  whose  ordinate  is  4. 
(See  §  28.)  Likewise,  the  complex 
number  -2+3  i  lies  at  (-2,  3) ;  the 
number  3  —  2  i  lies  at  (3,  —2) ;  and,  in 
general,  the  number  x+yi,  where  x 
and  y  are  any  (real)  numbers,  lies  at 
the  point  (x,  y).  Whenever  a  plane 
is  used  in  this  way  to  represent  com- 
plex numbers,  it  is  called  a  complex 
plane.  The  re-axis  is  called  the  axis 


. 

Y 

& 

1 

e 

f 

X4 

y 

8 

3 

1 

5- 

-4i 

-±5 

i 
\ 

.Q- 

/ 

/ 

/ 

,/ 

< 

/ 

/ 

\ 

I/ 

Ax 

S  0 

f  r« 

als 

0 

^ 

Sw 

X 

^ 

3-2i 

FIG.  39. 


of  reals  and  the  ?/-axis  is  called  the  axis  of  pure  imaginaries  (because 
the  pure  imaginary  part  of  each  complex  number  is  to  be  measured 
parallel  to  it).  The  straight  line  joining  any  point  x+iy  to  the 
origin  is  called  the  radius  vector  of  that  point,  or  number. 

Observe  that  from  this  point  of  view  all  real  numbers  become 
represented  by  the  points  on  the  axis  of  reals,  while  all  pure  imaginary 
numbers  become  represented  by  the  points  on  the  axis  of  pure 
imaginaries,  the  other  points  of  the  plane  being  then  taken  up  by 
what  are  properly  the  complex  numbers. 


CHAPTER  XIV 
SIMULTANEOUS    QUADRATIC   EQUATIONS 

I.   ONE  EQUATION  LINEAR  AND  THE  OTHER  QUADRATIC 

78.  Graphical  Solution.  In  Chapter  VI  we  have  seen  how 
we  may  determine  graphically  the  solution  of  two  linear 
equations  e&ch  of  which  contains  the  two  unknown  letters 
x,  y.  The  method  consists  in  first  drawing  the  graph  of  each 
equation,  then  observing  the  x  and  the  y  of  the  point  where 
the  two  graphs  intersect  (cut  each  other).  The  particular 
pair  of  values  (x,  y)  thus  obtained  constitutes  the  solution. 

We  often  meet  with  simultaneous  equations  which  are  not 
both  linear,  as  for  example  the  two  equations 

(1)  x-y  =  l, 

(2)  z2+2/2  =  25. 

Here  the  first  equation  is  linear  (§  26)  but  the  second  is  not. 
In  order  to  solve  them,  that  is  in  order  to  find  that  pair  (or 
pairs)  of  values  of  x  and  y  which  satisfy  both  equations,  we 
may  proceed  graphically  as  follows. 

The  graph  of  (1)  is  found  (as  in  §  29)  to  be  the  straight 
line  shown  in  Fig.  40. 

To  draw  the  graph  of  (2),  we  first  solve  this  equation  for  y. 
Thus  y2  =  25  -  x2.  Therefore 


(3)  2/=± 

122 


XIV,  §  78]         SIMULTANEOUS  QUADRATICS 


123 


We  now  work  out  a  table  of  pairs  values  of  x  and  y  that  will 
satisfy  (3),  that  is  we  give  x  various  values  .in  (3)  and  solve 
for  the  corresponding  y  value.  (Compare  §  60.)  The  result 
is  shown  below.  Observe  that  to  the  value  x  =  0  there  cor- 


Whenz  = 

0 

+  1 

+2 

+3 

+4 

+5 

then     y  = 

±V25 
±5 

W24 
±4.8 

±V21 

±4.5 

±VlQ 
±4 

±V9 
±3 

±  Vo 

0 

respond  the  £wo  values  T/  =  ±  5  ;  similarly  to  x  =  1  correspond 
the  two  values  y=  ±4.8  (approximately),  etc. 

Moreover,  if  we  assign  to  x  the  negative  value,  x=—l, 
we  find  in  the  same  way  that  corresponding  to  it  y  has  the 
two  values  y  =  ±4.8.  Likewise,  for  x  =  —  2  we  find  y  =  ±4.5, 
etc.,  the  values  of  y  for  any  negative  value  of  x  being  the 
same  each  time  as  for  the  corresponding  positive  value  of  x. 


FIG.  40. 

Plotting  all  the  points  (x,  y)  thus  obtained  and  drawing 
the  smooth  curve  through  them,  the  graph  is  as  shown  in 
Fig.  40.  This  curve  is  a  circle,  as  appears  more  and  more 
clearly  as  we  plot  more  and  more  points  (x,  y)  belonging  to 
the  equation  (3). 


124  SECOND   COURSE   IN  ALGEBRA      [XIV,  §  78 

NOTE.  The  form  of  (3)  shows  that  there  can  be  no  points  in  the 
graph  having  x  values  greater  than  5,  for  as  soon  as  x  exceeds  5  the 
expression  25  —  x2  becomes  negative  and  hence  V25  —  x2  becomes 
imaginary,  and  there  is  no  point  that  we  can  plot  corresponding  to 
such  a  result.  Similarly,  it  appears  from  (3)  that  x  cannot  take 
values  less  than  —5. 

Thus  the  graph  can  contain  no  points  lying  outside  the  circle 
already  drawn. 

Returning  now  to  the  problem  of  solving  (1)  and  (2),  we 
know  (§31)  that  wherever  the  one  graph  cuts  the  other  we 
shall  have  a  point  whose  x  and  y  form  a  solution  of  (1)  and 
(2),  that  is  we  shall  have  a  pair  of  values  (x,  y)  that  will 
satisfy  both  equations  at  once.  From  the  figure  it  appears 
that  there  are  in  the  present  case  two  such  points,  namely 
(z  =  4,  2/  =  3)  and  (x=-3,  y=-4).  Equations  (1)  and  (2) 
therefore  have  the  two  solutions  (z  =  4,  2/  =  3)  and  (x=—  3, 
y——  4).  Ans. 

CHECK.  For  the  solution  (x  =  4,  y=3)  we  have  x—  y=4  —  3  =  1, 
and  x2  +yz  =  16  +9  =  25,  as  required. 

For  the  solution  (x  =  —3,  y  =  —4)  we  have  x  -y  =  -3  -  (  -4)  =  1, 
and  z2 +?/2=  9 +16  =  25,  as  required. 

The  following  are  other  examples  of  the  graphical  study  of 
non-linear  simultaneous  equations. 

EXAMPLE  1.     Solve  the  system 
(4) 
(5) 

SOLUTION.  The  straight  line  representing  the  graph  of  (4)  is 
drawn  readily. 

To  obtain  the  graph  of  (5),  we  have  9  y2  =  100  —4  x2.     Hence 

y2  =£(100 -4  z2)  =|(25  -x2), 
and  therefore 
(6) 


XIV,  §  78]         SIMULTANEOUS   QUADRATICS 
Corresponding  to  (6),  we  find  the  following  table : 


125 


When  z= 

0 

+  1 

+2 

+3 

+4 

+5 

greater  than 
+5 

then  y= 

ifV/25 
±§(5) 
±3.3 

±3^/24 
±1(4.8) 
±3.2 

±|V/21 
±§(4.5) 
±3.0 

±|X/16 
±§(4) 
±2.6 

±|V^ 
±§(3) 

±2 

±1^0 
±0 
0 

imaginary 
imaginary 
imaginary 

For  any  negative  value  of  x,  the  y- values  are  the  same  as  for  the 
corresponding  positive  value  of  x.  (See  the  solution  of  (1 )  and  (2) .) 

The  graph  thus  obtained  for  (6),  or  (5),  is  an  oval  shaped  curve. 
It  belongs  to  a  general  class  of  curves  called  ellipses. 


\ 


FIG.  41. 

The  two  graphs  are  seen  to  intersect  at  the  points 
(x  =  4,  y  =2)  and  (x  =-5,y  =  0). 

Therefore  the  desired  solutions  of  (4)  and  (5)  are  (x=4,  y=2) 
and  (x=  —  5,  2/=0).     Ans. 

•  EXAMPLE  2.     Solve  the  system 

(7)  2x-y=-2, 

(8)  Z2/  =  4. 

SOLUTION.    The  graph  of  (7)  is  the  straight  line  shown  in  Fig.  42. 
To  obtain  the  graph  of  (8),  we  have 

(9)  y=±, 


126 


SECOND   COURSE   IN  ALGEBRA        [XIV,  §  78 


from  which  we  obtain  the  following  table  : 


When  x  = 

8 

7 

6 

5 

T 

4 

3 
t 

2 

1 

* 

i 

ST 

i 

T 

16 

i 

5" 

then    y  = 

i 

* 

t 

1 

2 

4 

8 

12 

20 

This  table  concerns  only  positive  values  of  x,  but  it  appears 
from  (9)  that  for  any  negative  value  of  x  the  appropriate  y  value 
is  the  negative  of  that  for  the  corresponding  positive  value  of  x. 

The  graph  thus  obtained  for  (9),  or  (8),  consists  of  two  open 
curves,  each  indefinitely  long,  situated  as  in  Fig.  42.  These  taken 
together  (that  is,  regarded  as  one  curve)  form  what  is  known  as  a 
hyperbola  (pronounced  hy-per'bo-la).  The  part  (branch)  lying  to 
the  right  of  the  i/-axis  corresponds  to  the  above  table,  while  the 
other  branch  corresponds  to  the  negative  x  values. 


/\ 


FIG.  42. 

The  two  graphs  are  seen  to  intersect  in  the  points  (a?  =  l,  y=4) 
and  (x=  -2,  y=  -2). 

Therefore  the  desired  solutions  of  (7)  and  (8)  are  (x  =  l,  j/=4) 
and  (x  =  —2,  y  =  —2).  Ans. 

NOTE.  Ellipses  and  hyperbolas  are  extensively  considered  in 
the  branch  of  mathematics  called  analytic  geometry  —  a  study  which 
may  be  pursued  after  a  course  in  plane  geometry  and  a  course  in 
algebra  equivalent  to  that  in  this  book.  It  is  usually  taught  during 
the  first  year  of  college  mathematics. 


XIV,  §  78]          SIMULTANEOUS   QUADRATICS 


127 


EXAMPLE  3.     Consider  graphically  the  system 

(10) 

(11) 

SOLUTION.  The  graph  of  (10)  is  found  in  the  usual  manner,  and 
is  represented  by  the  straight  line  in  Fig.  43.  The  graph  of  (11) 
has  already  been  worked  out  (see  discussion  of  (2)),  being  a  circle 
of  radius  5  with  center  at  the  origin.  The  peculiarity  to  be  es- 
pecially observed  here  is  that  these  two  graphs  do  not  intersect. 
This  means  (as  it  naturally  must)  that  there  are  no  real  solutions 
to  the  system  (10)  and  (11) ;  in  other  words,  the  only  possible  solu- 
tions are  imaginary. 


FIG.  43. 

Likewise,  whenever  any  two  graphs  fail  to  intersect,  we  may  be 
assured  at  once  that  the  only  solutions  their  equations  can  have 
are  imaginary.  The  system  (10)  and  (11)  and  other  such  systems 
will  be  considered  further  in  the  next  article. 

EXERCISES 

Draw  the  graphs  for  the  following  systems  and  use  your 
result  to  determine  the  solutions  whenever  they  are  real. 


4  z2+9  i/2  =  100. 
[HINT.     See  Example  1,  §  78. 


128  SECOND   COURSE    IN   ALGEBRA       [XIV,  §  78 

3.    J*-20=-l,  6 


\y  =  x*. 
[HINT.     See  Example  2,  §  78.] 

O. 

5~ q  „, n 
Jj  —  o  u  —  u. 

79.   Solution  by  Elimination.     Let  us  consider  again  the 
system  (1)  and  (2)  of  §  78. 
(1)  x-y=l, 

Instead  of  solving  this  system  graphically,  we  may  solve 
it  by  elimination,  that  is  by  the  process  employed  with  two 
linear  equations  in  §  33. 

Thus  we  have  from  (1) 

(3)  y  =  x-l. 

Substituting  this  value  of  y  in  (2),  thus  eliminating  y  from 
(2),  we  obtain 

or,  dividing  through  by  2, 

(4)  z2-z-12  =  0. 

Solving  (4)  by  formula  (§  56),  gives  as  the  two  roots 

1+7 


=  4, 


and 


(-l)-V(-l)«-4(l)(-12)_l-Vl+48  =  l-7^_q 
2  22 

When  x  has  the  first  of  these  values,  namely  4,  we  see 
from  (3)  that  y  must  have  the  value  ?/  =  4—  1,  or  3. 


XIV,  §  79]         SIMULTANEOUS   QUADRATICS  129 

Similarly,  when  x  takes  on  its  other  value,  namely  —  3,  we 
see  that  y  has  the  value  y=  —  3  —  1,  or  —4. 

The  solutions  of  the  system  (1)  and  (2)  are,  therefore, 
(^  =  4,  ^  =  3)  and  (x=  -3,  y=  -4).  Ans. 

Observe  that  these  results  agree  with  those  obtained 
graphically  for  (1)  and  (2)  in  §  78. 

Further  applications  of  this  method  are  made  in  the 
examples  that  follow. 

EXAMPLE  1.     Solve  the  system 

(5)  2x-\-y  =  4, 

(6)  z2+2/2  =  12. 
SOLUTION.     From  (5), 

(7)  y=±-2x. 
Substituting  this  expression  for  y  in  (6),  we  find 


or 

(8)  5z2-16z+4  =  0. 

The  two  roots  of  (8),  as  determined  by  formula  (§  56),  are 

=  -  ( - 16)  =fc  V  ( - 16)2  -4(5)  (4)  =  16  ±  V256  -80=  16  =*=  vT76 
2(5)  10  10 

_16^4VII^8=>=2Vn 

10  5 

The  first  of  these  values,  namely  x=   (8+2vTF)/5,  when  sub- 
stituted in  (7),  gives  as  its  corresponding  value  of  y, 
j,.4_ia+4vH_4-|V] 

The  second  value,  namely  x  =  (8—  2VTT)/5,  when  substituted 
in  (7),  gives  as  its  corresponding  value  of  y, 

16-4VT1     4+4VTT 


, 

5 
Hence  the  desired  solutions  are 


x    8+2VTT 

8-2VTI 

5            and 
4-4VH 

5        ' 

4+4^11 

y            5        ' 

y"          5 

130  SECOND   COURSE    IN  ALGEBRA        [XIV,  §  79 

To  obtain  the  approximate  values  of  the  numbers  thus  obtained 
we  have  VTI  =3.31662  (tables),  and  hence  the  above  solutions  re- 
duce to  the  forms 

x=  2.9266,  (x  =0.2734, 

y=-  1.8533,  \y  =  3.4533. 

These  are  the  solutions  of  the  system  (5),  (6),  correct  to  four  places 
of  decimals,  which  is  sufficient  for  ordinary  work. 

NOTE.  It  may  be  remarked  that,  while  the  graphical  method  of 
solution  described  in  §  78  is  very  instructive  in  showing  how  many 
solutions  a  given  system  will  have,  and  what  their  geometric 
significance  is,  it  does  not  usually  afford  a  ready  means  of  deter- 
mining the  exact  values  of  the  solutions.  This  is  illustrated  in  the 
example  just  solved,  where,  if  the  graphs  of  (5)  and  (6)  be  drawn, 
they  will  intersect  at  points  whose  x  and  y  contain  the  surd  Vll 
(as  the  above  solution  shows),  and  it  would  be  difficult  to  measure 
off  any  such  values  accurately  on  the  scale  of  the  diagram.  In 
fact,  it  would  be  practically  impossible  to  determine  graphically 
the  solutions  of  (5)  and  (6)  correct  to  three  or  even  two  places  of 
decimals,  yet  this  degree  of  approximation  was  easily  obtained  above 
by  the  method  of  elimination.  For  such  reasons,  it  is  preferable, 
whenever  one  is  concerned  only  with  finding  the  values  of  solutions, 
to  proceed  from  the  beginning  by  the  method  of  elimination. 

EXAMPLE  2.     Solve  the  system 

(9) 
(10) 

SOLUTION.     From  (9), 

(11)  ^  =  10-z. 

Substituting  this  expression  in  (10), 

x2  +  (100  -20  z  +z2)  =  25, 
or 

(12)  2  x2-  20x+75=0. 

Solving  (12)  by  formula,  we  find  its  solutions  to  be 

and 


Since  these  z-values  contain  the  square  root  of  the  negative 
quantity  -200,  they  are  imaginary  (§58).     The  ^-values  are  also 


XIV,  §  79]         SIMULTANEOUS   QUADRATICS 


131 


imaginary,  as  appears  by  substituting  the  z-values  just  found  into 
(9),  which  gives  the  results 


The  desired  solutions  of  the  systems  (9),  (10)  are  therefore 


x=- 


10 


-=-    and 


10 


10 


y  = 


10 


This  result  should  now  be  contrasted  with  what  we  saw  in 
Example  3  of  §  78  regarding  this  same  system  (9)  and  (10).  There 
we  found  graphically  that  the  solutions  must  be  imaginary  be- 
cause the  graphs  failed  to  intersect,  but  we  could  not  find  the  actual 
imaginary  numbers  which  form  the  solutions.  This  we  have  now 
been  able  to  do,  however,  by  the  method  of  elimination.  The 
method  of  elimination  enables  one  to  determine  imaginary  as  well 
as  real  solutions  in  all  similar  cases. 

EXERCISES 

Solve  each  of  the  following  systems  by  the  method  of  elimi- 
nation, and,  in  case  surds  are  present,  find  each  solution 
correct  to  two  places  of  decimals  by  use  of  the  tables. 

f    x+y=-l, 


1. 


2. 


f    x-2y=-l, 


6. 

7. 


f    x-2y  =  2, 

O«  i  <>      *       A          9  Of 


8    i 

9. 


x-2y  =  3. 

*-2y*=-S, 

x-2y=-3. 

f  3o;2-a-5     = 


15' 


xy=-Q. 

2x+y  =  2, 
xy=-Q. 


f 

10.     I  x-\-y    x  —  y     6' 


132 


SECOND    COURSE   IN  ALGEBRA       [XIV,  §  80 


II.  NEITHER  EQUATION  LINEAR 

*80.  Two  Quadratic  Equations.  In  each  of  the  systems 
considered  in  §§  78,  79  one  of  the  two  given  equations  was 
linear.  However,  the  same  methods  of  solving  may  often 
be  employed  in  case  neither  equation  is  linear.  In  such  cases 
four  solutions  may  be  present  instead  of  two. 


EXAMPLE  1.     Solve  the  system 


(1) 
(2) 


z2-?2=15. 


SOLUTION.  Here  only  xz  and  y2  appear  and  we  begin  by  finding 
their  values.  Thus,  multiplying  (2)  through  by  16  and  adding  the 
result  to  (1),  we  eliminate  y2  and  find  that  25  z2  =400,  or 

(3)  z2  =  16. 


Substituting  this  value  of  x  in  (2),  we  find 


(4) 


From  (3)  and  (4)  we  now  obtain 


(5) 


x= 


and     y  ==*=!. 


Forming  all  the  pairs  of  values 
x,  y  that  can  come  from  (5),  we 
obtain  as  our  desired  solutions 

(x  =  4,  y  =  —  1) ;  and 
(x=  —  4,  y=  —  1).     Ans. 

CHECK.  Each  of  these  pairs  of 
values  of  x  and  y  is  immediately 
seen  to  satisfy  both  (1)  and  (2). 
Let  the  pupil  thus  check  each  pair. 
When  considered  graphically, 
equation  (1)  gives  rise  to  an 
ellipse  (compare  §  78,  Ex.  1),  while 

(2)  gives  a  hyperbola  situated  as  shown  in  the  diagram.  These 
two  curves  intersect  in  four  points  which  correspond  to  the  four 
solutions  just  obtained. 


FIG.  44. 


XIV,  §  80]        SIMULTANEOUS   QUADRATICS 


133 


EXAMPLE  2.     Solve  the  system 


-l2. 


(7) 

(8)  xy 

SOLUTION.     Here  we  cannot  proceed  as  in  Example  1  because 
we  cannot  find  readily  the  values  of  x2  and  y*.     But  if  we  multiply 

(8)  by  2  and  add  the  result  to  (7),  we  obtain 

(9)  x2  +2  xy  +y*  =  l. 

Taking  the  square  root  of  both  members  of  (9)  gives 


(10) 

Similarly,  multiplying  (8) 
by  2  and  subtracting  the  result 
from  (7), 

x2  -  2  xy  +?/2  =  49, 
and  hence 

(11)  X-y=±7. 

Taking  account  of  the  two 
choices  of  sign  in  (10)  and 
(11),  we  see  that  they  give  rise 
to  the  four  simple  (linear) 
systems 

(a)  x+y  =  l,  x-y=7; 

(6)  x+y=-l,  x-y=7; 

(c)  x+y  =  l,  x-y=-7; 

(d)  x+y=-l,x-y=-7', 

Thus  we  have  replaced  the  original  system  (7)  and  (8)  by  the 
four  simple  systems  (a),  (6),  (c),  and  (d),  each  of  which  may  be 
immediately  solved  by  elimination,  as  in  §  §  33,  34.  Since  the  solu- 
tions of  (a),  (6),  (c),  (d)  are  respectively  (x=4,  y=  -3),  (x=3, 
y  =  —4),  (x  =  -3,  y  =4),  and  (x  =  —4,  y  =3),  we  conclude  that  these 
are  the  desired  solutions  of  (7)  and  (8).  Ans. 

The  graphical  significance  of  these  solutions  is  shown  in  Fig.  45, 
where  the  circle  z2+?/2  =  25  is  cut  by  the  hyperbola  xy=  —  12  in 
four  points  that  correspond  to  the  four  solutions  just  found. 

CHECK.  That  these  four  solutions  each  satisfy  (7)  and  (8) 
appears  at  once  by  trial. 


FIG.  45. 


134  SECOND   COURSE   IN   ALGEBRA        [XIV,  §  80 

While  no  general  rule  can  be  stated  for  solving  two  equations 
neither  of  which  is  linear,  the  following  observation  may  be 
made.  Unless  the  equations  can  be  solved  readily  for  x2 
and  y2  (as  in  Example  1),  the  system  should  first  be  carefully 
examined  with  a  view  to  making  such  combinations  of  the 
given  equations  as  will  yield  one  or  more  new  systems  each 
of  which  can  be  solved  (as  in  Example  2)  by  methods  already 
familiar.  All  solutions  obtained  in  this  way  should  be 
checked,  since  false  combinations  of  the  x-  and  ^-values  are 
frequently  made  by  beginners  when  the  work  becomes  at  all 
complicated. 

EXERCISES 

Solve  each  of  the  following  systems,  and  draw  a  diagram 
for  each  of  the  first  three  to  show  the  geometric  meaning  of 
your  solutions. 

=  25,  f 

=  7.  \ 

f 


[HINT  FOR  Ex.  4.  First  add,  then  subtract  the  two  equations, 
thus  showing  that  the  given  system  is  equivalent  to  two  others, 
namely 


Now  solve  each  of  these  systems  as  in  §  80.] 
z2+92/2  =  85, 


, 

(     x- 


y+2xy=-2Q. 


CARDAN 
(Girolamo  Cardan,  1501-1576) 

An  equation  of  the  third  degree  (cubic  equation)  has  three  roots  and 
these  can  be  found  only  by  methods  which  are  more  powerful  than  those 
employed  in  the  study  of  quadratics.  Cardan  was  the  first  to  obtain  and 
publish  a  method  for  solving  such  equations.  His  methods  are  also  sufficient 
to  solve  any  pair  of  simultaneous  quadratics,  but  are  too  advanced  to  be 
given  in  this  book. 


XIV,  §  81J         SIMULTANEOUS   QUADRATICS  135 

*81.  Systems  Having  Special  Forms.  The  systems  of  equations 
considered  in  §^  79,  80  illustrate  the  usual  and  more  simple  types 
such  as  one  commonly  meets  in  practice.  It  is  possible,  however, 
to  solve  more  complicated  systems  provided  they  are  of  certain 
prescribed  forms.  We  shall  here  consider  only  two  such  type  forms. 

I.   When  one  (or  both)  of  the  given  equations  is  of  the  form 


where  the  coefficients  o,  6,  c  are  such  that  the  expression  ax2+bxy+cy2 
can  be  factored  into  two  linear  factors. 
EXAMPLE.     Solve  the  system 

(1)  x*+2x-y  =  7, 

(2)  x2-xy-2y*=0. 

SOLUTION.     Here  we  see  that  (2)  is  of  the  form  mentioned  above, 
since  x2—xy—2  yz  can  be  factored  (as  in  §12  (e)}  into  (x—  2  y)(x+y}. 

(2)  may  thus  be  written  in  the  form 

(3)  (x-2y)(x+y)=Q. 
It  follows  (§  52)  that  either 

x-2y=Q,      or      x+y=Q. 

Hence  the  system  (1),  (2)  may  be  replaced  by  the  two  following 
systems  : 


and 

fz2+2o;-2/  =  7, 
I  x+y=0. 

Each  of  these  two  systems  may  now  be  solved  as  in  §  80,  and  we 
thus  find  that  the  solutions  of  the  first  system  are 

(x  =  2,   y  =  l)  and  (x=-%,y=—  J) 
while  the  solutions  of  the  second  system  are 


and 


_ 

y=i(3+V37). 


The  desired  solutions  of  (1)  and  (2)  consist,  therefore,  of  these  four 
solutions  just  obtained.     Ans. 


136  SECOND   COURSE    IN   ALGEBRA       [XIV,  §  81 

II.    When  both  the  given  equations  are  of  the  form 

axz+bxy+cy*=d, 

where  a,  6,  c,  and  d  have  any  given  values  (0  included). 
EXAMPLE.     Solve  the  system 

(4)  x*-xy+y*  =  3, 

(5)  x*+2xy=5. 

SOLUTION.     Let  v  stand  for  the  ratio  x/y  ;    that  is,  let  us  set 


y 

Then 

(6)  x=vy. 
Substituting  in  (4), 

(7)  0V 
Substituting  in  (5), 

(8) 

Solving  (7)  for  y*, 

(9) 


Solving  (8)  for  y2, 

(10)  ?/2 

Equating  the  values  of  yz  given  by  (9)  and  (10), 


v2+2v     v2- 
Clearing  of  fractions, 
(11)  2vz-llv+5  = 

Solving  (11)  by  formula  (§56), 

^H=faVl21-40 

4  44 

Therefore   v  =  5,  or  v  =%.     Substituting  5  for  v  in  (9),  or  (10), 
Hence 

=      i    or L 

Vf  C         V7 
Substituting  ^  for  v  in  (9)  or  (10),  yz  =4.     Hence  y  =  +2,  or  -2. 


XIV,  §  82]         SIMULTANEOUS   QUADRATICS  137 

The  only  values  that  y  can  have  are,  therefore,  1/V?,  -1/V?, 
2,  and  -2. 

Since  x_=vy  (see  (6)),  the  value  of  x  to  go  with  y  =  l/V7  is 
x=5(l/v/7)=5/V7.  _Similarly,  when  y=  -1/V?  we  have 
x  =5(  —  I/VT)  =  —  5/V?.  Likewise,  when  ?/=2  (in  which  case 
v  =  -^,  as  shown  on  p.  136)  then  z  =^  •  2  =  1,  and  when  y  —  —  2,  then 
z=i(-2)  =  -l. 

Therefore  the  only  solutions  which  the  system  (4),  (5)  can  have 
are(x=5/V7,  y  =  l/V7)  ;  (x  =  -5/S/7,  »=  -1/V?)  ;  (x  =  1,  y  =  2)  ; 
OT=—  1,  i/=—  2);  and  it  is  easily  seen  by  checking  that  each  of  these 
is  a  solution.  Ans. 

82.  Conclusion.  Every  system  of  equations  considered 
in  this  chapter  has  been  such  that  we  could  solve  it  by  finally 
solving  one  or  more  simple  quadratic  equations.  We  have 
examined  only  special  types,  however,  and  the  student  should 
not  conclude  that  all  pairs  of  simultaneous  quadratics  can  be 
solved  so  simply. 

MISCELLANEOUS   EXERCISES 

Solve  the  following  simultaneous  quadratics.  The  star 
(*)  indicates  that  the  exercise  depends  upon  §  81. 


3. 


*   y 

2--        = 


[HINT  TO  Ex.  4.     First  eliminate  xy  between  the  two  equa- 
tions so  as  to  obtain  a  linear  equation  between  x  and  y.] 

z2+2z-2/  =  5,  I 

2  z2-3  x+2  y  =  S.  \ 


, 

lx2-i/2  =  9.  i 

[HINT  TO  Ex.  6.     Divide  the  first  equation  by  the  second.] 


138 


SECOND   COURSE    IN   ALGEBRA     -  [XIV,  §  82 


10 


xy-y=l2. 


:15. 


APPLIED   PROBLEMS 

In  working  the  following  problems,  let  x  and  y  represent  the  two 
unknown  quantities,  then  form  two  simultaneous  equations  and  solve 
them.  If  surds  occur,  find  their  approximate  values  by  the  tables. 

1.  The  sum  of  two  numbers  is  13  and  the  difference  of 
their  squares  is  91.     Find  the  numbers. 

2.  A  piece  of  wire  48  inches  long  is  bent  into  the  form 
of  a  right  triangle  whose  hypotenuse  is  20  inches  long.     What 
are  the  lengths  of  the  sides?     (See  Ex.  14  (d),  p.  6.) 

3.  If  it  takes  26  rods  of  fence  to  inclose  a  rectangular  garden 
containing  ^  of  an  acre,  what  are  the  length  and  breadth  ? 

4.  Figure  46  shows  two  circles  just 
touching  (tangent  to)  each  other,  the 
smaller  one  being  outside  the  larger  one. 
If  their  combined  area  is  15f  square  feet 
and  the  distance  CC'  between  the  two 
centers  is  3  feet,  find  the  radius  of  each 
circle. 


FIG.  46. 


5.  Work  Ex.  4  in  case  the  circles  touch 
on  the  inside  of  the  larger  one,  taking  the 
shaded  area  to  be  110  square  feet  and 
CC'  to  be  5  feet. 


FIG.  47. 


XIV,  §  82]         SIMULTANEOUS   QUADRATICS  139 

6.  Do  positive  integers  exist  differing  by  5  and  such  that 
the  difference  of  their  squares  is  45  ?     If  so,  find  them. 

7.  Answer  the  question  in    Ex.    6    in    case    the    differ- 
ence of  the  squares  is  taken  to  be  10,  other  conditions  re- 
maining the  same. 

8.  The  area  of  a  certain  triangle  is  160  square  feet,  and 
its  altitude  is  twice  as  long  as  its  base.     Find,  correct  to 
three  decimal  places,  the  base  and  the  altitude.     (See  Ex. 
14  (a),  p.  6.) 

9.  The  area  of  a  rectangular  lot  is  2400  square  feet,  and 
the  diagonal  across  it  measures  100  feet.     Find,  correct  to 
three  decimal  places,  the  length  and  breadth. 

10.  The  mean  proportional  between  two  numbers  is  2 
and  the  sum  of  their  squares  is  10.     What  are  the  numbers  ? 
(See  Ex.  6,  page  80.) 

11.  The  dimensions  of  a  rectangle  are  5  feet  by  2  feet. 
Find  the  amounts  (correct  to  two  decimal  places)  by  which 
each  dimension  must  be  changed,  and  how,  in  order  that 
both  the  area  and  the  perimeter  shall  be  doubled. 

12.  Two  men  working  together  can  complete  a  piece  of 
work  in  6  days.     If  it  would  take  one  man  5  days  longer 
than  the  other  to  do  the  work  alone,  in  how  many  days 
can  each  do  it  alone?     (Compare  Ex.  9,  p.  56.) 

13.  The  fore  wheel  of  a  carriage  makes  28  revolutions 
more  than  the  rear  wheel  in  going  560  yards,  but  if  the 
circumference  of  each  wheel  be  increased  by  2  feet,   the 
difference   would    be   only   20    revolutions.     What    is    the 
circumference  of  each  wheel? 

14.  A  sum  of  money  on  interest  for  one  year  at  a  certain 
rate  brought  $7.50  interest.     If  the  rate  had  been  1%  less 
and  the  principal  $25  more,  the  interest  would  have  been  the 
same.     Find  the  principal  and  the  rate. 


140  SECOND   COURSE    IN   ALGEBRA       [XIV,  §  82 

15.  A  man  traveled  30  miles.  If  his  rate  had  been  5 
miles  more  per  hour,  he  could  have  made  the  journey  in  1 
hour  less  time.  Find  his  time  and  rate. 

16.  Figure  48  shows  a  circle  within 
which  a  diameter  A B  has  been  drawn. 
At  a  certain  point  P  on  AB  the  perpen- 
B  dicular  PG  measures  4  inches,  while  at 
the  point  Q,  which  is  1  unit  from  P, 
the  perpendicular  QF  measures  3  inches. 
How  long  is  the  diameter  AB1 

[HINT.  Let  x  =  AP,y  =  PB.  Find  x  and  y,  using  the  fact  stated 
in  Ex.  6,  page  80,  then  take  their  sum.] 

17.  Show  that  the  formulas  for  the  length  I  and  the 
width  w  of  the  rectangle  whose  perimeter  is  a  and  whose 
area  is  b  are 


18.  Find  the  formulas  for  the  radii  of  two  circles  in  order 
that  the  difference  of  the  areas  of  the  circles  shall  be  d  and 
the  sum  of  their  circumferences  shall  be  s. 

and  fi..«?=i£i    Ans. 


4  7TS  4  TTS 

19.  Find  two  fractions  whose  sum  is  f  ,  and  whose  differ- 
ence is  equal  to  their  product. 

20.  The  diagonal  and  the  longer  side  of  a  rectangle  are 
together  five  times  the  shorter  side,  and  the  longer  side 
exceeds  the  shorter  side  by  35  yards.     What  is  the  area  of 
the  rectangle? 


CHAPTER  XV 
PROGRESSIONS 

I.  ARITHMETIC  PROGRESSION 

83.  Definition.  An  arithmetic  progression  is  a  sequence 
of  numbers,  called  terms,  each  of  which  is  derived  from  the 
preceding  by  adding  to  it  a  fixed  amount,  called  the  common 
difference.  An  arithmetic  progression  is  denoted  by  the 
abbreviation  A.  P. 

Thus  1,  3,  5,  7,  •••  is  an  A.  P.  Each  term  is  derived  from  the 
preceding  by  adding  2,  which  is  therefore  the  common  difference. 
The  dots  indicate  that  the  sequence  may  be  extended  as  far  as  one 
pleases.  Thus  the  first  term  after  7  would  be  9,  the  next  one  would 
be  11,  etc. 

Again,  5,  1,  —3,  —  7,  — 11,  •••  is  an  A.  P.  Here  the  common 
difference  is  —4. 

EXERCISES 

Determine  which  of  the  following  are  arithmetic  progres- 
sions ;  determine  the  common  difference  and  the  next  two 
terms  of  each  of  the  arithmetic  progressions. 

1.  3,  6,  9,  12,  ••-.        4.  30,  25,  20,  15,  10,  •••. 

2.  3,  5,  8,  12,  ....        6.  -1,  -1^,  -2,  -2£,  •••. 

3.  6,  4,  2,  0,  -2,  -4,  •••.   6.  a,  2  a,  4  a,  5  a,  •••. 

7.  a,  a+3,  a+6,  a+9,  •••. 

8.  a,  a-\-d,  a+2  d,  a+3  d,  a+4  d,  •••. 

9.  x-4y,  x-2y,  x-y,  •••. 

10.  3x+3y,  Qx+2y,  9  x+y,  •••. 

[HINT.  The  common  difference  may  always  be  determined  by 
subtracting  any  term  from  the  term  immediately  preceding.] 

141 


142  SECOND   COURSE    IN   ALGEBRA        [XV,  §  83 

11.   Write  the  first  five,  terms  of  the  A.  P.  in  which 
(a)  The  first  term  is  5  and  the  common  difference  is  2. 
(6)  The  first  term  is  —3  and  the  common  difference  1. 
(c)   The  first  term  is  3  a  and  the  common  difference  is  —  b. 

84.  The  nth  Term  of  an  Arithmetic  Progression.  From 
the  definition  (§  83)  it  follows  that  every  arithmetic  progres- 
sion is  of  the  form  a,  a+d,  a+2  d,  a+3  d,  a+4  d,  •••.  Here 
a  is  the  first  term  and  d  the  common  difference. 

Observe  that  the  coefficient  of  d  in  any  one  term  is  1  less 
than  the  number  of  that  term.  Thus  2  is  the  coefficient  of 
d  in  the  third  term;  3  is  the  coefficient  of  d  in  the  fourth 
term,  etc.  Therefore  the  coefficient  of  d  in  the  nth  term 
must  be  (n—  1).  Hence,  if  we  let  I  stand  for  the  nth  term, 
we  have  the  formula 


EXAMPLE.     Find  the  llth  term  of  the  A.  P.  1,  3,  5,  7,  •  •  •. 

SOLUTION.     We  have  a  =  l,  d  =2,  n  =  ll,  1=  ? 

The  formula  gives  Z  =  a  +  (n-l)d  =  l+10X2  =  1+20  =  21.     Ans. 

EXERCISES 

1.  Find  the  llth  term  of  3,  6,  9,  12,  •••. 

2.  Find  the  13th  term  of  6,  10,  14,  18,  •••. 

3.  Find  the  20th  term  of  4,  2,  0,  -2,  -4,  •••. 

%4.    Find  the  15th  term  of  -1,  -1£,  -2,  -2J,  •••. 

5.  Find  the  10th  term  of  x-y,  2  x  —  2  y,  3  x-3  y,  •  •  •. 

6.  When  a  small  heavy  body  (like  a  bullet)  drops  to  the 
ground  it  passes  over  16.1  feet  the  first  second,  3  times  as  far 
the  second  second,  5  times  as  far  the  third  second,  etc.     How 
far  does  it  go  in  the  12th  second? 

7.  If  you  save  5  cents  during  the  first  week  in  January, 
10  cents  the  second  week,  15  cents  the  third  week  and  so  on, 
how  much  will  you  save  during  the  last  week  of  the  year? 


XV,  §  85]  PROGRESSIONS  143 

8.  What  term  of  the  progression  2,  6,  10,  14,  •••  is  equal 
to  98? 

[HINT,     a  =  2,  d=4,  n=?,  /=98.] 

9.  What  term  of  3,  7,  11,  15,  •••  is  equal  to  59? 

10.  The  first  term  of  an  A.  P.  is  8  and  the  14th  term  is  47. 
What  is  the  common  difference? 

85.  The  Sum  of  the  First  n  Terms  of  an  Arithmetic  Pro- 
gression. Let  a  be  the  first  term  of  an  A.  P.,  d  the  common 
difference,  I  the  nth  term.  Then  the  sum  of  the  first  n 
terms,  which  we  will  call  S,  is 

(1)  s  =  a+(a+d)  +  (a+2d)  +  (a+Zd)+''.+  (l-d)+l. 
This  value  for  S  may  be  written  in  a  very  much  simpler 

form,  as  we  shall  now  show. 

Write  the  terms  of  (1)  in  their  reverse  order.     This  gives 

(2)  S  =  l+(l-d)  +  (l-2d)  +  (l-3d)+---+(a+d)+a. 

Now  add  (1)  and  (2),  noting  the  cancellation  of  d  with  —dt 
2  d  with  -2  d,  etc.     The  result  is 


or  2S  =  n(a+l). 

Therefore  S  =  -(a+/). 

2 

This  is  the  simple  form  for  S  mentioned  above.  If  we  re- 
place I  by  its  value  a+(n—l)d  (§84),  this  result  takes  the 
form 


2 

EXAMPLE.     Find  the  sum  of  the  first  12  terms  of  the  A.  P. 
2,6,  10,  14,  —. 

SOLUTION,     a  =  2,   d=4,   n  =  12. 

Therefore,  by  the  second  form  for  S  in  §  85,  we  have 

4}  =  6(4+44}  =  6X48  =288.     Ans. 


144  SECOND   COURSE    IN  ALGEBRA         [XV,  §  85 

EXERCISES 

Find  the  sum  of  each  of  the  following  arithmetic  progres- 
sions. 

1.  The  first  ten  terms  of  3,  6,  9,  12,  •••. 

2.  The  first  fifteen  terms  of  -2,  0,  2,  4,  •••. 

3.  The  first  thirteen  terms  of  1,  3^,  6,  ••-. 

4.  The  first  ten  terms  of  1,  -1,  -3,  -5,  •••. 

5.  The  first  n  terms  of  1,  8,  15,  •••. 

6.  How  many  strokes  does  a  common  clock,  striking 
hours,  make  in  12  hours? 

7.  A  body  falls  16.1  feet  the  first  second,  3  times  as  far 
the  second  second,  5  times  as  far  the  third  second,  etc.     How 
far  does  it  fall  during  the  first  12  seconds? 

8.  Find  the  sum  of  all  odd  integers,  beginning  with  1  and 
ending  with  99. 

9.  If  you  save  5  cents  during  the  first  week  in  January, 
10  cents  during  the  second  week,  15  cents  the  third  week, 
and  so  on,  how  much  will  you  save  in  a  year? 

10.  The  first  term  of  an  A.  P.  is  4  and  the  10th  term  is  31. 
What  is  the  sum  of  the  10  terms? 

[HINT,     a  =  4,  ft  =  10,  Z=31.     Now  use  the  first  of  the  formulas 
in  §  85.] 

11.  The  first  term  of  an  A.  P.  is  £  and  the  12th  term  is 
11-J-.     What  is  the  sum  of  the  12  terms? 

12.  Figure  49  shows   a    series   of  16 
dotted  lines  which  are  equally   distant 
from  each  other.     If  the  highest  one  is 
6  inches  long  and  the  lowest  one  is  3  feet 

long,  what  is  the  sum  of  all  their  lengths?  Fl°-  49- 

[HINT.     The  lines  form  an' A.  P.  since  their  lengths  increase  uni- 
formly.] 


XV,  §  85] 


PROGRESSIONS 


145 


13.  The  rungs  of  a  ladder  diminish  uniformly  from  2  feet 
4  inches  long  at  the  base  to  1  foot  3  inches  long. at  the  top. 
If  there  are  24  rungs,  what  is  the  total  length  of  wood  in 
them? 


14.  Find  the  sum  of  the  circumferences 
of  10  concentric  circles  if  the  radius  of  the 
innermost  one  is  ^  inch  and  the  radius  of 
the  outermost  one  is  4  inches,  it  being  under- 
stood that  the  circles  are  equally  spaced  from 
each  other. 


FIG.  50. 


15.  Figure  51  shows  a  coil  of  rope  in  the 
ordinary  circular  form,  containing  12  com- 
plete turns,  or  layers.  If  the  ength  of  the 
innermost  turn  is  4  inches  and  the  length  of 
the  outermost  turn  is  37  inches,  how  long  is 
the  rope? 


FIG.  51. 


[HINT.  Regard  each  turn  as  a  circle,  thus  neglecting  the  slight 
effect  due  to  the  overlapping  at  the  beginning  of  each  turn  after  the 
first.] 

16.  If  in  Fig.  51  the  length  of  the  innermost  turn  is  a 
inches  and  that  of  the  outermost  turn  is  b 
inches,  and  the  number  of  turns  is  n,  what 
represents  the  total  length  of  the  rope? 


17.  A  small  rope  is  wound  tightly  round  a 
cone,  the  number  of  complete  turns  being  24. 
Upon  unwinding  the  rope  from  the  top,  the 
lengths  of  the  first  and  second  turns  are 
found  to  be  2J  inches  and  3^  inches  respec- 
tively. How  long  (approximately)  is  the  rope? 


FIG.  62. 


146  SECOND   COURSE    IN   ALGEBRA         [XV,  §  86 

86.   Arithmetic  Means.     The  terms  of  an  arithmetic  pro- 
gression lying  between  any  two  given  terms  are  called  the 
arithmetic  means  between  those  two  terms. 

Thus,  the  three  arithmetic  means  between  1  and  9  are  3,  5,  7, 
since  1,  3,  5,  7,  9  form  an  arithmetic  progression. 

Whenever  a  single  term  is  inserted  in  this  way  between  two 
numbers,  it  is  briefly  called  the  arithmetic  mean  of  those  two 
numbers. 

Thus,  the  arithmetic  mean  of  2  and  10  is  6,  because  2,  6,  10  form 
an  arithmetic'  progression. 

A  formula  for  the  arithmetic  mean  of  any  two  numbers, 
as  a  and  b,  is  easily  obtained.  Thus,  if  x  is  the  mean,  then 
a,  x,  b  forms  an  A.  P.  Therefore,  we  must  have  x  —  a  =  b—x. 
Solving  for  x,  this  gives 

x_a±b 

2 

Thus  we  have  the  following  theorem :  The  arithmetic 
mean  of  two  numbers  is  equal  to  half  their  sum. 

NOTE.  The  arithmetic  mean  of  two  numbers  is  also  called  their 
average. 

EXAMPLE.     Insert  five  arithmetic  means  between  3  and  33. 

SOLUTION.  We  are  to  have  an  A.  P.  of  7  terms  in  which  a  =3, 
I  =33,  and  n  =7.  We  begin  by  finding  d.  Thus, 

Z  =  a  +  (n-l)d  (§  84)  so  that  33=3+6d.     Solving,  d=5. 

The  progression  is  therefore  3,  8,  13,  18,  23,  28,  33  and  hence 
the  desired  means  are  8,  13,  18,  23,  28.  Ans. 

EXERCISES 

1.  Insert  three  arithmetic  means  between  7  and  23. 

2.  Insert  four  arithmetic  means  between  —5  and  10. 

3.  Insert  seven  arithmetic  means  between  ^  and  25f . 

4.  What  is  the  arithmetic  mean  of  8  and  30  ? 
6.  What  is  the  arithmetic  mean  of  J  and  — -J-? 


XV,  §  86] 


PROGRESSIONS 


147 


/ 


FIG.  53. 


B 


6.  Show  that  the  first  formula  for  S  obtained  in  §  85 
may  be  stated  as  follows :   "  The  sum  of  n  terms  of  an  arith- 
metic progression  is  equal  to  n  multiplied  by  the  arithmetic 
mean  of  the  first  and  nth  terms." 

7.  A  BCD  is  any  trapezoid  (that  is,  any  four-sided  figure 
having  its  bases  A  B  and  DC  parallel  to  each  other).     The 
line   EF,   called    the  median,   joins    the 

middle    point    of   the    side   AD    to   the 

middle  point  of  the  side  BC,  and  it  is 

shown  in  geometry  that    the  length  of    ^ 

this  line  EF  will  always  be  the  arith-    A 

metic  mean  of  the  lengths  of  the  bases 

AB  and  CD.     Hence  answer  the  following  questions. 

(a)  If  the  bases  are  10  inches  long  and  2  inches  long, 
respectively,  what  is  the  length  of  the  median? 

(b)  If  the  lower  base  is  14  inches  long  and  the  median  8 
inches  long, -how  long  is  the  upper  base? 

(c)  If  the  upper  base  is  3  feet  long  and  the  median  4  feet 
long,  how  long  is  the  lower  base? 

(d)  If  the  bases  are  a  inches  long  and  b  inches  long,  re- 
spectively, what  represents  the  length  of  the  median? 

8.  The  figure  shows  the  frustum  of  a  cone  and  the  frus- 
tum of  a  pyramid,  and  in  each  case  the  "  mid-section  "  has 


FIG.  54. 


been  drawn  in  (that  is,  the  section  made  by  a  plane  which 
passes  midway  between  the  bases  AB  and  BC).  It  is  shown 
in  solid  geometry  that  in  all  such  cases  the  perimeter  of  the 


148  SECOND   COURSE    IN   ALGEBRA         [XV,  §  86 

mid-section  will  always  be  the  arithmetic  mean  of  the 
perimeters  of  the  two  bases.  Hence  answer  the  following 
questions. 

(a)  If  the  perimeters  of  the  two  bases  are  30  inches  and  10 
inches  respectively,  what  is  the  perimeter  of  the  mid-section  ? 

(6)  In  the  frustum  of  a  cone,  the  radius  of  the  upper  base 
is  2  inches  and  that  of  the  lower  base  8  inches.  What  is  the 
perimeter  of  the  mid-section? 

87.  The  Five  Elements  of  an  Arithmetic  Progression.     In 

any  arithmetic  progression  there  are  the  five  elements,  a,  d, 
I,  n,  Sj  defined  in  §§  84,  85.  If  any  three  of  these  are  given, 
we  can  always  find  the  other  two  by  means  of  the  formulas 
in  §§  84,  85. 

EXAMPLE  1.  Given  a  =-•§•,  ft  =  30,  S  =  2l±.  Find  d 
and  I. 

SOLUTION.    From  §  85,  we  have     S  =(a  +Z). 


Hence  21i  =  15( 

Solving,  I  =  lii.     Now,  I  =  a  +  (n  -  l)d.     Hence  1£J  =  -i  +29  d. 

Solving,  d=^. 

EXAMPLE  2.     Given  a  =  3,  d  =  4,  £  =  300.     Find  n  and  I 

SOLUTION.     S=%{2  a  +  (n-l)d}.     Hence  300=S6  +  O-1)-  4}. 
2  2i 

Therefore 

600  =  n{4n+2};  4  n2+2  n-600=0;  2n2+n-300=0. 
Solving  the  last  (quadratic)  equation  by  formula  (§56),  gives 


or  _ 


•4  44 

Since  n  is  the  number  of  terms  and  therefore  a  positive  integer, 
it  follows  that  n  =  12.     (See  Hint  to  Ex.  3,  p.  89.) 

To  find  Z,  we  now  use  the  formula  I  =  a  +  (n  —  l)d.     Thus 

1=3+11  -4=3+44  =  47. 


XV,  §  87]  PROGRESSIONS  149 

EXERCISES 

1.  Given  a  =  3,  n  =  25,  £  =  675,  find  d  and /. 

2.  Given  a=  -9,  n  =  23,  Z  =  57,  find  d  and  S. 

3.  Given  £  =  275,  1  =  4:5,  n  =  ll,  find  a  and  d. 

4.  Find  w  and  d  when  a  =  -  5,  I  =  15,  S  =  105. 

5.  Find  a  and  n  when  1  =  1,  d=%,  £=-20. 

6.  How  many  terms  are  there  in  the  arithmetic  pro- 
gression 2,  6,  10,  —70? 

7.  Given  a,  I,  and  n,  derive  a  formula  for  d. 

8.  Given  a,  d,  and  Z,  derive  a  formula  for  n. 

9.  Given  a,  n,  and  S,  derive  a  formula  for  Z. 

10.  Given  d,  I,  and  S,  derive  a  formula  for  a. 

11.  Find  an  A.  P.  of  14  terms  having  13  for  its  6th  term 
and  25  for  its  10th  term. 

12.  Find  an  A.  P.  of  16  terms  such  that  the  sum  of  the  6th, 
7th  and  8th  terms  is  —  16J-,  and  the  sum  of  its  last  two  terms 
is  -28. 

13.  Find  three  integers  in  arithmetic  progression  such  that 
their  sum  is  24  and  their  product  384. 

14.  The  figure  represents  one  of  the  four 
sides  of  a  steel  tower  such  as  is  commonly  seen 
at  wireless  telegraph  stations.  It  is  desired  to 
make  one  of  these  towers  so  that  each  girder, 
such  as  A  B,  will  be  2  feet  longer  than  the  one 
just  above  it,  as  CD.  How  many  girders  will 
the  tower  have  (counting  all  four  sides)  in  case 
the  total  amount  of  girder  steel  used  is  to  be 

only  864  feet  and  the  lowest  girders  are  each  to 
FIG.  55.      haye  a  lenth  of  20  feet? 


150  SECOND   COURSE    IN   ALGEBRA         [XV,  §  88 

II.   GEOMETRIC   PROGRESSION 

88.  Definitions.  A  geometric  progression  is  a  sequence 
of  numbers,  called  terms,  each  of  which  is  derived  from  the 
preceding  by  multiplying  it  by  a  fixed  amount,  called  the 
common  ratio.  A  geometric  progression  is  denoted  by  the 
abbreviation  G.  P. 

Thus  2,  4,  8,  16,  32,  •  •  •  is  a  G.  P.  Each  term  is  derived  from  the 
preceding  by  multiplying  it  by  2,  which  is  therefore  the  common 
ratio. 

Again,  10,  —5,  +-JJ-,  •—•§-,  •••  is  a  G.  P.  whose  common  ratio  is  — -g- 
The  next  two  terms  are  +f ,  -y5¥. 

EXERCISES 

Determine  which  of  the  following  are  geometric  progres- 
sions, and  find  the  common  ratio  and  the  next  two  terms  of 
each  geometric  progression. 

1.  3,6,  12,24,48,  •••. 

2.  4,  12,  48,  75,---. 

3.  *,i,i,A,  •;••• 

4.  -1,2,  -4,8,  -16,".. 

5.  a,  a2,  a3,  a4,  ••-. 

6.  2x,  4z3,  8z5,  16  z7,  ••«. 

7.  a,  ar,  ar2,  or3,  or4,  •••. 

8.  a,  aV2,  a3r4,  a  r6,  •••. 

9.  (0+6),  (a+6)3,  (a+6)5,  (o+&)7,  -. 
in  m2  m4  ra6  m8 

n3'  n4'  n^  n«J  ' 

11.   Write  the  first  five  terms  of  the  G.  P.  in  which 
(a)  The  first  term  is  4  and  the  common  ratio  is  4. 
(6)  The  first  term  is  —3  and  the  common  ratio  is  —2. 
(c)   The  first  term  is  a  and  the  common  ratio  is  r. 


XV,  §  89]  PROGRESSIONS  151 

89.  The  nth  Term  of  a  Geometric  Progression.  From  the 
definition  in  §  88  it  follows  that  every  geometric  progression 

is  of  the  form 

a,  ar,  ar2,  ar3,  ar4,  ••-. 

Here  a  is  the  first  term,  and  r  the  common  ratio. 

Observe  that  the  exponent  of  r  in  any  one  term  is  1  less 
than  the  number  of  that  term.  Thus  2  is  the  exponent  of  r 
in  the  third  term  ;  3  is  the  exponent  of  r  in  the  fourth  term, 
etc.  Therefore,  the  exponent  of  r  in  the  nth  term  must  be 
(n  —  l).  Hence,  if  we  let  Z  stand  for  the  nth  term,  we  have 
the  formula 


EXAMPLE.     Find  the  7th  term  of  the  G.  P.  6,  4,  f, 

SOLUTION.     We  have  a  =6,  r=§,  n=7,  l=? 

The  formula  gives  I  =  or»-i  =  6  X  (|)  '  =2  X3  x|°  =|  =  Ans. 


EXERCISES 

1.  Find  the  ninth  term  of  2,  4,  8,  16,  •••. 

2.  Find  the  eighth  term  of  ^,  ^,  1,  •••. 

3.  Find  the  ninth  term  of  —  1,  2,  —  4,  8,  •••. 

4.  Find  the  tenth  term  of  4,  2,  1,  -J,  •••. 

5.  Find  the  eighth  term  of  -*-,  £,  f  , 

6.  Find  the  eleventh  term  of  ax,  a?x2,  a3z3,  a4#4,  •••. 

7.  Find  the  tenth  term  of  2,  A/2,  1,  ••-. 

8.  What  term  of  the  G.  P.  3,  6,  12,  24  is  equal  to  384? 

9.  What  term  of  the  progression  6,  4,  f  is  equal  to  |4  ? 
10.  For  every  person  there  has  lived  two  parents,  four 

grandparents,  eight  great  grandparents,  etc.  How  many 
ancestors  does  a  person  have  belonging  to  the  7th  genera- 
tion before  himself,  assuming  that  there  is  no  duplication? 
Answer  also  for  the  10th  generation. 


152  SECOND   COURSE   IN  ALGEBRA         [XV,  §  89 

11.  If  you  save  50  cents  during  the  first  three  months  of 
the  year  and  double  the  amount  of  your  savings  every  three 
months  afterward,  how  much  will  you  save  during  the  last 
three  months  of  the  second  year  ? 

12.  From  a  grain  of  corn  there  grew  a  stalk  that  produced 
an  ear  of  100  grains.     These  grains  were  planted  and  each 
produced  an  ear  of  100  grains.     This  was  repeated  until 
there  were  5  harvestings.     If  75  ears  of  corn  make  a  bushel, 
how  many  bushels  were  there  the  fifth  year? 

90.  The  Sum  of  the  First  n  Terms  of  a  Geometric  Progres- 
sion. Let  a  be  the  first  term  of  a  geometric  progression,  r 
the  common  ratio,  I  the  nth  term.  Then  the  sum  of  the  first 
n  terms,  which  we  will  call  S}  is 

(1)  /S  =  a+ar+ar2+arH  -----  \-arn-2+arn~l. 

This  value  for  S  may  be  written  in  a  very  much  simpler 
form,  as  we  shall  now  show. 

Multiply  both  members  of  (1)  by  r.     This  gives 

(2)  rS  =  ar+ar2+arH  -----  \-arn~l-\-arn. 

Now  subtract  equation  (2)  from  equation  (1),  noting  the 
cancelation  of  terms.  This  gives 

S-rS  =  a-arn. 
Solving  this  equation  for  S  gives 


l-r 

This  is  the  simple  form  for  S  mentioned  above. 

It  is  to  be  observed  also  that  since  I  =  arn~l,  we  may  write 
rl  =  arn.  Putting  this  value  of  arn  into  the  form  just  found 
for  S,  we  obtain  as  a  second  expression  for  S  the  following 

formula. 

fl-r? 


XV,  §  90]  PROGRESSIONS  153 

EXAMPLE.     Find  the  sum  of  the  first  six  terms  of  the  G.  P. 
3,  6,  9,  12,  .-. 

SOLUTION,     a  =3,  r  =  2,  n=6.     To  find  S. 
<y     a -ar*_3-3.26_3 -3-64 _3 -192 _  -189 _1pn       , 

=T^7=:~r=2~=   ^r~    ^r    ^r= 

EXERCISES 

Find  the  sum  of  the  first 

1.  Eight  terms  of  2,  4,  8,  •••. 

2.  Six  terms  of  1,  5,  25,  •••. 

3.  Five  terms  of  1,  !£,  2£,  — . 

4.  Six  terms  of  2,  — f ,  f,  •-. 

5.  Ten  terms  of  -£,  i,  -£,  •••. 

6.  Six  terms  of  1,  2  a,  4  a2,  •••. 

7.  Ten  terms  of  1,  a2,  a4,  .... 

8.  What  is  the  sum  of  the  series  3,  6,  12,  ••-,  384? 

9.  What  is  the  sum  of  the  series  8,  4,  2,  •••,  -j^? 

10.  Find  the  sum  of  the  first  ten  powers  of  2. 

11.  Find  the  sum  of  the  first  seven  powers  of  3. 

12.  A  series  of  five  squares  are  drawn  such  that  a  side  of 
the  second  one  is  twice  as  long  as  a  side  of  the  first  one,  a 
side  of  the  third  one  is  twice  as  long  as  a  side  of  the  second, 
etc.     If  a  side  of  the  first  one  is  2  inches  long,  find  (by  §  90) 
the  sum  of  the  areas  of  all  the  squares. 

13.  What  is  the'  combined  volume  of  five  spheres  if  the 
radius  of  the  first  one  is  16  inches,  the  radius  of  the  second 
one  is  half  that  of  the  first  one,  the  radius  of  the  third  one  is 
half  that  of  the  second  one,  and  so  on  to  the  fifth  one  ?     (See 
Ex.  14  (e),  p.  6.) 

14.  Half  the  air  in  a  certain  corked  empty  jug  is  removed 
by  each  stroke  of  an  air  pump.     What  fraction  of  the  original 
volume  of  air  has  been  removed  by  the  end  of  the  seventh 
stroke  ? 


154  SECOND   COURSE    IN   ALGEBRA        [XV,  §  90 

HISTORICAL  NOTE.  It  is  related  that  when  Sessa,  the  inventor 
of  chess,  presented  his  game  to  Scheran,  an  Indian  prince,  the  latter 
asked  him  to  name  his  reward.  Sessa  begged  that  the  prince  would 
give  him  1  grain  of  wheat  for  the  first  square  of  the  chess  board,  2 
for  the  second,  4  for  the  third,  8  for  the  fourth,  and  so  on  to  the  sixty- 
fourth.  Delighted  with  the  inventor's  modesty,  the  prince  ordered 
his  ministers  to  make  immediate  payment.  The  number  of  grains 
of  wheat  thus  called  for  was  (see  §  90) 

1-1.2"=2*-1=2<U     1 

1-2  1 

But  the  value  of  264  is  the  enormous  number  18,446,744,073,709, 
551,616,  so  the  number  of  grains  of  wheat  owing  was  but  1  less  than 
this.  This  amount  is  greater  than  the  world's  annual  supply  at  pres- 
ent. History  does  not  relate  how  the  claim  was  settled.  (From 
Godfrey  and  Siddons'  Elementary  Algebra,  Vol.  II,  pp.  336,  337.) 

91.  Geometric  Means.  The  terms  of  a  geometric  progres- 
sion lying  between  any  two  given  terms  are  called  the  geo- 
metric means  of  those  two  terms. 

Thus  the  three  geometric  means  of  2  and  32  are  4,  8,  16,  since 
2,  4,  8, 16,  32 form  a  geometric  progression. 

Whenever  a  single  term  is  inserted  in  this  way  between  two 
numbers,  it  is  briefly  called  the  geometric  mean  of  those  two 
numbers. 

Thus  the  geometric  mean  of  2  and  32  is  8,  since  2,  8,  32  form  a 
geometric  progression. 

A  formula  for  the  geometric  mean  of  any  two  numbers, 
as  a  and  b,  is  easily  obtained.  Thus,  if  x  is  the  mean,  then  a, 
x,  b  forms  a  G.  P.  Therefore  we  must  have  x/a  =  b/x.  Solv- 
ing, we  have  z2  =  ab,  and  hence 

x  =  Vab. 

Thus,  we  have  the  following  theorem  :  The  geometric  mean 
of  two  numbers  is  equal  to  the  square  root  of  their  product. 

NOTE.  The  geometric  mean  of  two  numbers  is  thus  the  same  as 
their  mean  proportional.  See  Ex.  6,  p.  80. 


XV,  §  91] 


PROGRESSIONS 


155 


EXAMPLE.     Insert  four  geometric  means  between  3  and  96. 

SOLUTION.     We  are  to  have  a  G.  P.  of  six  terms  in  which  a  =3, 
I  =  96,  and  n  =  6.     We  begin  by  finding  r.     Thus  • 

I  =  arn-l(§  89)  so  that  96  =  3  •  r5,  or  r5  =  32.     Hence  r  =  2. 

The  progression  is  therefore  3,  6,  12,  24,  48,  96,  and  hence  the 
desired  means  are  6,  12,  24,  48.     Ans. 

EXERCISES 

1.  Insert  four  geometric  means  between  2  and  486. 

2.  Insert  three  geometric  means  between  1  and  625. 

3.  Insert  five  geometric  means  between  4J  and  y^. 

4.  What  is  the  geometric  mean  of  2  and  18? 

5.  What  is  the  geometric  mean  of  8  and  50  ? 

6.  What  is  the  geometric  mean  of  \  and  3f . 

7.  Find,  correct  to  four  decimal  places,  the  geometric 
mean  of  6  and  27,  using  the  tables  of  square  roots. 

8.  Find,  correct  to  four  decimal  places,  the  geometric 
mean  of  2J  and  3^. 

9.  Insert  two  geometric  means  between  5  and  9,  express- 
ing each  correct  to  four  decimal  places. 

10.  Show  that  the  number  of  units  in  a  side  of  the  square 
is  the  geometric  mean  of  the  number  of  units  in  the  two  un- 
equal sides  of  a  rectangle  that  has  the  same  area. 

11.  Figure  56  shows  a  square  within  which 
is  placed  (in  any  manner)  another  square 
whose  side  is  half  as  long  as  that  of  the 
first  square.     Show  that  the  area  between 
the  squares  is  equa1  to  three  halves  of  the 
mean  proportional  between  the'  areas  of  the 

squares  themselves.  FIG.  56. 

12.  Show  that  the  result  stated  in  Ex.  11  holds  true  also 
in  the  case  of  the  area  between  two  circles,  the  smaller  circle 
lying  within  the  larger  and  having  its  radius  half  as  long  as 
that  of  the  larger  circle.     Draw  a  figure. 


156  SECOND   COURSE    IN   ALGEBRA         [XV,  §  92 

92.  Infinite  Geometric  Progression.  Consider  the  geo- 
metric progression 

(i)  !,*,*,*,  A,  -• 

Here  o  =  l,  r=%,  and  hence,  by  §  90,  the  sum  of  n  terms  is 
-ar 
1-r 

Now,  if  the  value  selected  for  n  is  very  large,  the  expres- 
sion (l/2)n,  which  here  appears,  is  very  small,  being  the  frac- 
tion £  multiplied  into  itself  n  times.  In  fact,  as  n  is  selected 
larger  and  larger,  this  expression  (l/2)w  comes  to  be  as  small 
as  we  please,  so  that  the  value  for  S,  as  given  above,  comes  as 
near  as  we  please  to 

1-0 


which  is  the  same  as  2.     So  we  say  that  2  is  the  sum  to  in- 
finity of  the  geometric  progression  above,  meaning  thereby 
simply  that  as  we  sum  up  the  terms,  taking  more  and  more 
of  them,  we  come  as  near  as  we  please  to  2. 
The  meaning  of  this  result  is  seen  in  the  figure  below. 


FIG.  67. 

Here,  beginning  at  the  point  marked  0,  we  first  measure 
off  1  unit  of  length,  then,  continuing  to  the  right,  we  measure 
off  %  unit,  then  -i  unit,  then  -J  unit,  etc.,  each  time  going  to 
the  right  just  one  half  the  amount  we  went  the  time  before. 
As  this  is  kept  up  indefinitely,  we  evidently  come  as  near  as 
we  please  to  the  point  marked  2,  which  is  2  units  from  0. 
This  corresponds  exactly  to  what  we  are  doing  when  we  add 
more  and  more  of  the  terms  of  the  given  progression 


XV,  §  92]  PROGRESSIONS  157 

A  progression  like  the  one  just  considered,  in  which  the 
value  of  n  is  not  stated  but  may  be  taken  as  large  as  one 
pleases,  is  called  an  infinite  geometric  progression. 

Having  thus  considered  the  sum  to  infinity  of  the  special 
infinite  geometric  progression  (1),  let  us  now  suppose  that  we 
have  any  infinite  geometric  progression,  as 

a,  ar,  ar2,  ar3,  •••, 

and  (as  before)  that  r  has  some  value  numerically  (§1)  less 
than  1.  Then  the  sum  of  the  first  n  terms  is 

o     a  —  arn 

=T^7' 

and,  as  n  is  taken  larger  and  larger,  the  expression  rn  which 
appears  here  becomes  as  small  as  we  please,  since  we  have 
supposed  r  to  be  less  than  1.  •  Hence,  as  n  increases  indefi- 
nitely, the  value  of  S  comes  as  near  as  we  please  to 

a-a  -0 
1-r   ' 
or 

a 
1-r' 

We  have  therefore  the  following  theorem:  The  sum  to 
infinity  of  any  geometric  progression  whose  common  ratio  r 
is  numerically  less  than  1  is  given  by  the  formula 


1-r 

EXAMPLE.     Find  the  sum  to  infinity  of  the  progression 

3>  1>  -g-j  ^j  -fr,  '"• 

SOLUTION,     a  =3,  r  =  ^.     Since  r  is  numerically  less  than  1,  we 
have  by  the  formula  of  §  92, 

<        a          3        39., 


158 


SECOND   COURSE   IN  ALGEBRA         [XV,  §  92 
EXERCISES 


Find  the  sum  to  infinity  of  each  of  the  following  progres- 
sions, and  state  in  each  case  what  your  answer  means,  draw- 
ing a  diagram  similar  to  Fig.  57  to  illustrate. 

2        Q      3         3          3 
•        O,    -I".    T'B't    ~a~Ti  . 


[HINT.     r=— i  and   hence   is   numerically  less   than   1.     The 
formula  of  §  92  therefore  applies.] 

4.    5,  .5,  .05,  .005,  •••. 

6.  1  —  x+x2  —  x3+  •••whenx=f. 

7.  V2,  1,4='  I   .., 


I    2    _2\/2  4 

1 3'  3vr  9'  '"* 


10.  A  pendulum  starts  at  A  and  swings 
to  B,  then  it  swings  back  as  far  as  C, 
then  forward  as  far  as  D,  etc.  If  the 
first  swing  (that  is,  the  circular  arc  from 
A  to  B)  is  6  inches  long  and  each  suc- 
ceeding swing  is  five  sixths  as  long  as  the 
one  just  preceding  it,  how  far  will  the 
pendulum  bob  travel  before  coming  to 
FIG.  58.  rest? 

11.   At  what  time  after  3  o'clock  do  the  hands  of  a  watch 
pass  each  other? 

[HINT.     We  may  look  at  this  as  follows :    The  large  (minute) 
hand  first  moves  down  to  where  the  small  (hour)  hand  is  at  the  be- 


XV,  §  93]  PROGRESSIONS  159 

ginning,  that  is  through  15  of  the  minute  spaces  along  the  dial. 

Meanwhile  the  small  hand  advances  T^  as  far  or  |f  of  a  minute  space. 

This  brings  the  small  hand  to  the  position  indicated  by  the  dotted 

line  in  the  figure.     The  large  hand  next  passes 

over  this  ^f  of  a  minute  space.     Meanwhile 

the  small  hand  again  advances  ^  as  far,  which 

is  -£f±  of  a  minute  space.     The  large  hand  next 

covers  this  ^f^  of  a  minute  space,  but  the  small 

hand  meanwhile  advances  ^  as  far,  or  yyf-g- 

of  a  minute  space,  etc.    Thus,  the  successive 

moves  of  the  large  hand,  counting  from  the  first 

one,  form   the  G.   P.   15,  if,  ^,  T^,  .... 

The  sum  of  this  to  infinity  will  be  the  total  distance  passed  over 

by  the  large  hand  before  the  hands  pass.] 


93.  Variable.  Limit.  We  have  seen  (§  92),  in  connection 
with  the  geometric  progression  1,  -g-,  -J-,  -J-,  ••'•,  that  the  sum  of 
its  first  n  terms  is  a  quantity  which,  as  n  increases  indefinitely, 
comes  and  remains  as  near  as  we  please  to  the  exact  value  2. 
The  usual  way  of  stating  this  is  to  say  that  as  n  increases, 
the  sum  of  the  first  n  terms  approaches  2  as  a  limit.  The  sum 
of  the  first  n  terms  is  here  called  a  variable  since  it  varies,  or 
changes,  in  the  discussion.  A  similar  remark  applies  to  all 
the  infinite  geometric  progressions  which  we  have  consid- 
ered. In  every  case  the  sum  to  infinity  is  the  limit  which 
the  sum  of  the  first  n  terms,  considered  as  a  variable  quantity, 
is  approaching. 

NOTE.  It  may  be  asked  whether  the  sum  of  the  first  n  terms  of 
the  G.  P.  1,  \,  \,  -|,  •••  could  ever  actually  reach  its  limit  2.  The 
answer  is  that  it  may  or  it  may  not,  depending  upon  circumstances. 
Thus,  if  we  think  of  the  terms,  beginning  with  the  second,  as  being 
added  on  at  the  rate  of  one  a  minute  we  could  never  reach  the  end  of 
the  adding  process,  since  the  number  of  the  terms  is  inexhaustible  and 
hence  the  minutes  required  would  have  no  end.  In  other  words, 
the  sum  of  the  first  n  terms  could  never  reach  its  limit  on  this  plan. 


160  SECOND   COURSE    IN   ALGEBRA         [XV,  §  93 

But  suppose  that  instead  of  this  we  were  to  add  on  the  terms  with 
increasing  speed  as  we  went  forward.  For  example,  suppose  we 
added  on  the  ^  in  ^  a  minute,  then  the  -J  in  \  of  a  minute,  then  the 
i  in  -1-  of  a  minute,  etc.  On  this  plan  we  would  actually  reach  the 
limit  2  in  2  minutes  of  time.  Here  the  constantly  increasing  speed 
of  the  adding  process  exactly  counterbalances  the  fact  that  we  have 
an  indefinitely  large  number  of  terms  to  add,  with  the  result  that  we 
reach  the  end  of  the  process  in  the  definite  time  of  2  minutes.  This 
idea  is  practically  illustrated  in  Ex.  11,  p.  159,  where  the  hands  of  the 
watch  would  never  pass  each  other  at  all  except  for  the  fact  that  the 
successive  moves  of  the  large  hand,  which  constitute  the  terms  of 
the  progression  15,  ^-|,  -^£±,  xrls"' '  *  *  are  added  on  in  less  and  less  time 
as  the  process  goes  on,  each  being  added  on  in  ^  the  time  occupied 
by  the  one  just  before  it. 

The  question  of  whether  a  variable  can  reach  its  limit  is  inti- 
mately connected  with  the  famous  problem  considered  by  the 
Schoolmen  in  the  Middle  Ages  and  known  as  the  problem  of  Achilles 
and  the  tortoise.  In  this  problem,  Achilles,  who  was  a  celebrated 
runner  and  athlete,  starts  out  from  some  point,  as  A,  to  overtake 
a  tortoise  which  is  at  some  point,  as  T,  the  tortoise  being  famous  for 
the  slow  rate  at  which  it  crawls  along.  Both  start  at  the  same  in- 
stant and  go  in  the  same  direction,  as  indicated  in  the  figure. 


A  T 

FIG.  60. 

Achilles  soon  arrives  at  the  point  T,  from  which  the  tortoise  started, 
but  in  the  meantime  the  tortoise  has  gone  some  distance  ahead. 
Achilles  now  covers  this  last  distance,  but  this  leaves  the  tortoise 
still  ahead,  having  again  gained  some  additional  distance.  This 
continues  indefinitely.  How,  therefore,  can  Achilles  ever  overtake 
the  tortoise?  The  Schoolmen  never  quite  answered  this  question 
satisfactorily  to  themselves.  The  secret  of  the  difficulty  lies  in  the 
fact  that,  as  in  the  other  problems  mentioned  above,  the  successive 
moves  which  Achilles  makes  are  done  in  shorter  and  shorter  inter- 
vals of  time,  with  the  result  that,  although  the  number  of  moves 
necessary  is  indefinitely  great,  they  can  all  be  accomplished  in  a 
definite  time. 


XV,  §  94]  PROGRESSIONS  161 

94.  Repeating  Decimals.  If  we  express  the  fraction  £| 
decimally  by  dividing  12  by  33  in  the  usual  way,  we  find  that 
the  quotient  is  .363636  •••,  the  dots  indicating  that  the  divi- 
sion process  never  stops  (or  is  never  exact)  but  leads  to  a 
never-ending  decimal.  However,  the  figures  appearing  in 
this  decimal  are  seen  to  repeat  themselves  in  a  regular  order, 
since  they  are  made  up  of  36  repeated  again  and  again. 
Such  a  decimal  is  called  a  repeating  decimal.  More  generally, 
a  repeating  decimal  is  one  in  which  the  figures  repeat  them- 
selves after  a  certain  point.  Thus,  .12343434  •  •  -,  1.653653653 
•••,  are  repeating  decimals. 

Let  us  now  turn  the  question  around.  Thus,  suppose 
that  a  certain  repeating  decimal  is  given,  as  for  example 
.272727  •••,  and  let  us  ask  what  fraction  when  divided  out 
gives  this  decimal.  This  kind  of  question  is  usually  too 
difficult  to  answer  in  arithmetic,  but  it  can  be  easily  answered 
as  follows  by  use  of  the  formula  in  §  92. 

Thus  the  decimal  .272727  •  •  •  may  be  written  in  the  form 

100  o  o  ~i~  ioooooo~f"   *"• 


This  is  an  infinite  geometric  progression  in  which  a  = 
r=TJTr.     The  sum  of  this  progression  to  infinity  must  be  the 
value  of  the  given  decimal.     Hence,  the  desired  value  is 

=  27      100  =  27=3      A 
100      99     99     11 

This  answer  may  be  checked  by  dividing  3  by  11,  the  re- 
sult being  .272727  •  •  -,  which  is  the  given  decimal. 

NOTE.  It  is  shown  in  higher  mathematics  that  every  rational 
fraction  in  its  lowest  terms  (that  is,  every  number  of  the  form  a/6, 
where  a  and  6  are  integers  prime  to  each  other)  gives  rise  when 
divided  out  to  a  never-ending,  repeating  decimal,  while  every  irra- 
tional number  (such  as  V2)  gives  rise  when  expressed  decimally  to 
a  never-ending  non-repeating  decimal. 
M 


162 


SECOND    COURSE    IN   ALGEBRA         [XV,  §  94 


EXERCISES 


Find  the  values  of  the  following  repeating  decimals  and 
check  your  answer  for  each  of  the  first  six. 


1.    .414141  •••. 
4.    .3414141  -.. 


2.    .898989  •••. 


3.    .543543543 


SOLUTION.     .3414141  •  •  •  =  .3  +  .0414141  •  •  • 

=  .3  +^(.414141-) 


6.    .6535353  - 

6.  1.212121  • 

7.  3.2151515 


_  3  _,_'!  ^  41  ^100 
~10+10X100X99 
=  A_i_^L=338  =  169      A 
10    990    990    495' 

8.  5.032032032 

9.  6.008008008 
10.  34.5767676  • 


MISCELLANEOUS    PROBLEMS 

I.   ARITHMETIC   PROGRESSION 

1.  What  will  be  the  cost  of  digging  a  20-foot  well  if  the 
digging  costs  50  cents  for  the  first  foot  and  increases  by  25 
cents  for  each  succeeding  foot  ? 

2.  Fifty-five  logs  are  to  be  piled  so  that  the  top  layer 
shall  consist  of  1  log,  the  next  layer  of  2  logs,  the  next  layer 
of  3  logs,  etc.      How  many  logs  will  lie  on  the   bottom 
layer  ? 

3.  In  a  potato  race  30  potatoes  are  placed  at  the  dis- 
tances 6  feet,  9  feet,  12  feet,  etc.,  from  a  basket.     A  player 
starts  from  the  basket,  picks  up  the  potatoes  and  carries  them, 
one  at  a  time,  to  the  basket.     How  far  does  he  go  altogether 
in  doing  this? 


XV,  §  94]  PROGRESSIONS  163 

4.  A  row  of  numbers  in  arithmetic  progression  is  written 
down  and  afterwards  all  erased  except  the  7th  and  the  12th, 
which  are  found  to  be  — 10  and  15,  respectively.     What  was 
the  20th  number? 

5.  If  your  father  gives  you  as  many  dimes  on  each  of 
your  birthdays  as  you  are  years  old  on  that  day,  how  old 
will  you  be  when  the  total  amount  he  has  given  you  in  this 
way  amounts  to  $12? 

6.  How  many   arithmetic  means   must  be  inserted  be- 
tween the  numbers  4  and  25  in  order  that  their  sum  may 
amount  to  87  ? 

7.  Prove  that  equal  multiples  of  the  terms  of  an  arith- 
metic progression  are  in  arithmetic  progression. 

8.  Prove  that  the  sum  of  n  consecutive  odd  integers,  be- 
ginning with  1,  is  n  . 

9.  The  sum  of  three  numbers  in  arithmetic  progression 
is  30  and  the  sum  of  their  squares  is  462.     What  are  the 
numbers  ? 

[HINT.     The  numbers  may  be  represented  as  x—  y,  x,  x+y. 
Form  two  equations  and  solve  for  x  and  y.} 

10.  If  a  person  saves  $20  the  first  month  and  $10  each 
month  thereafter,  how  long  before  his  total  savings  will 
amount  to  $1700? 

11.  Divide  80  into  four  parts  which  are  in  arithmetic 
progression  and  which  are  such  that  the  product  of  the  first 
and  fourth  is  to  the  product  of  the  second  and  third  as  2:3. 

12.  Find  the  sum  of  the  first  40  terms  of  an  A.P.  in  which 
the  ninth  term  is  136  and  the  sum  of  the  first  nineteen  terms 
is  2527. 

13.  lid  =  2,  n  =  21,  and  5  =  147,  find  a  and  I 

14.  Show  that  if,  in  any  A. P.,  the  values  of  d,  I,  and  S  are 
given,  then  the  formula  for  a  is 


164  SECOND   COURSE    IN   ALGEBRA        [XV,  §  94 

II.   GEOMETRIC  PROGRESSION 

15.  A  wheel  in  a  certain  piece  of  machinery  is  making 
32  revolutions  per  second  when  the  steam  is  turned  off  and 
the  wheel  begins  to  slow  down,  making  one  half  as  many 
revolutions  each  second  as  it  did  the  preceding  second.     How 
long  before  it  will  be  making  only  2  revolutions  per  second? 

16.  Show  that  if  a  principal  of  $p  be  invested  at  r  % 
compound  interest,  the  sum  of  money  accumulating  at  the 
ends  of  successive  years  will  form  a  geometric  progression, 
while  if  the  investment  be  made  at  simple  interest,  the  sums 
accumulating  will  form  an  arithmetic  progression. 

17.  From  a  cask  of  vinegar  ^  the  contents  is  drawn  off 
and  the  cask  then  filled  by  pouring  in  water.     Show  that  if 
this  is  done  6  times,  the  cask  will  then  contain  more  than 
90%  water. 

[HINT.  Call  the  original  amount  of  vinegar  1,  then  express  (as 
a  proper  fraction)  the  amount  of  water  in  the  cask  after  the  first 
refilling,  second  refilling,  etc.] 

18.  A  set  of  concentric  circles  is  drawn,  each  having  a 
radius  half  that  of  the  circle  just  outside  it.     Show  that  the 
limit  toward  which  the  sum  of  their  circumferences  is  ap- 
proaching is  equal  to  twice  the  circumference  of  the  largest 
circle. 

19.  A  dipper  when  hung  on  a  wall  often  swings  back  and 
forth  for  a  time,  the  swings  gradually  dying  out.     If  the  first 
swing  occupies  1  second,  and  each  succeeding  swing  takes 
.9  as  long  as  the  one  before  it,  how  long  before  the  dipper 
comes  to  rest? 

20.  It  is  found  by  experiment  that  the  number  of  bacteria 
in  a  sample  of  milk  doubles  every  3  hours.     What  increase 
will  there  be  in  24  hours,  assuming  that  all  outside  conditions 
remain  the  same? 


XV,  §  94]  PROGRESSIONS  165 

21.  In  Fig.  61  a  series  of  ordinates  equally  spaced  from 
each  other  has  been  drawn,  the  first  one  being  laid  off  1  unit 
long,  the  second  one  being  laid  off  equal 
to  the  first  one  increased  by  %  its  length, 
the  third  being  equal  to  the  second  in- 
creased by  ^  its  length,  etc.  Show  that 
these  ordinates  represent  the  successive 
terms  of  the  G.  P.  whose  first  term  is  1 
and  whose  common  ratio  is  14-.  In  this 


sense,  the  figure  may  be  called  the  dia- 
gram for  the  G.  P.  in  which  a  =  l,  r=l-J. 

22.  Draw  the  diagram  for  the  G.  P.  in  which 

(a)  a  =  l,  r  =  H;          (6)  a  =  2,  r  =  l£;          (c)  a  =  4,  r  =  J. 

[HINT.     Use  8  ordinates  only,  spacing  them  at  any  convenient 
but  equal  distance  apart.] 

23.  Prove  that  any  series  of  numbers  formed  by  writing 
down  the  reciprocals  of  the  successive  terms  of  a  geometric 
progression  is  itself  a  geometric  progression. 

24.  Three  numbers  whose  sum  is  24  are  in  arithmetic  pro- 
gression, but  if  3, 4,  and  7  be  added  to  them  respectively,  the 
results  form  a  geometric  progression.     Find  the  numbers. 

25.  If  a  series  of  numbers  are  in  G.  P.,  are  their  squares 
likewise  in  G.  P.  ?    Answer  the  same  for  their  cubes ;  also 
for  their  square  roots  and  their  cube  roots. 

Answer  the  same  questions  for  an  A.  P. 

[HiNT.     See  that  your  reasoning  is  general,  that  is,  do  not  base  it 
upon  an  examination  of  some  special  cases.] 


CHAPTER  XVI 
RATIO   AND   PROPORTION 

95.  Ratio.  The  quotient  of  one  number  divided  by  another 
of  the  same  kind  is  called  their  ratio. 

Thus  the  ratio  of  6  inches  to  3  inches  is  |-,  or  2 ;  the  ratio  of  5  Ib. 
to  3  Ib.  is  |>  etc.  Note  that  in  each  of  these  cases  the  ratio  is  simply 
a  fraction  of  the  kind  studied  in  arithmetic. 

The  first  number,   or  dividend,  is  called  the  antecedent; 
the  second  number,  or  divisor,  is  called  the  consequent. 
Thus,  in  the  ratio  -J,  the  antecedent  is  3  and  the  consequent  is  4. 

EXERCISES 

1.  What  is  the  ratio  of.  10  yards  to  2  yards?  of  7  yards 
to  3  yards? 

2.  State  (as  a  fraction  in  its  simplest  form)  the  value  of 
each  of  the  following  ratios. 

(a)  5  to  25.      (c)  i  to  |.     (e)  1  to  3.          (g)   18  xz  to  4  z2. 
(6)   16  to  12.    (d)  2toi.     (/)  3  a  to  6  b.     (h)  x*-y*tox-y. 

3.  State  which  is  the  antecedent  and  which  the  conse- 
quent in  each  of  the  parts  of  Ex.  2. 

4.  What  is  the  ratio  of  10  inches  to  2  feet? 

[HINT.  First  reduce  the  2  feet  to  inches  so  that  we  may  com- 
pare like  numbers,  that  is  numbers  measured  in  the  same  unit.] 

5.  The  dimensions  of  a  certain  grain  bin  are  3  feet  by 
6  feet  by  7  feet.     What  is  the  ratio  of  its  cubical  contents  to 
that  of  a  bin  whose  dimensions  are  3  feet  6  inches  by  5  feet 
by  1£  yards? 

166 


XVI,  §  96]  RATIO   AND   PROPORTION  167 

6.  If  one  square  has  its  sides  each  twice  as  long  as  the 
sides  of  another  square,  what  is  the  ratio  of  the  area  of  the 
first  square  to  that  of  the  second? 

[HINT.     Let  a  =a  side  of  the  smaller  square.] 

7.  If  one  cube  has  its  edges  each  twice  as  long  as  the  edges 
of  another  cube,  what  is  the  ratio  of  the  volume  of  the  first 
cube  to  that  of  the  second  ? 

8.  Show  that   if   a   cylinder   and   a   cone 
have  the  same  circular  base  and  the  same 
height,  the  ratio  of  the  volume  of  the  cylinder 
to  that  of  the  cone  is  3  :  1. 

FIG.  62. 

9.  When  we  sharpen  a  lead  pencil  a  cer- 
tain part  of  the  cylindrical  lead  is  exposed.     What  part  of 
the  exposed  lead  is  cut  off  when  a  smooth  conical  point  is 
made? 

96.  Proportion.  A  proportion  is  an  expression  of  equality 
between  two  ratios,  or  fractions. 

For  example,  since  j-  is  the  same  as  f-,  we  have  the  proportion  ^  =  -|. 

Likewise,  we  may  write  f  =  £,  f=if>  -f  =  -f,  etc. ;  hence 
all  these  are  true  proportions.  But  f  =  -J-  is  not  a  proportion  since 
these  two  fractions  are  unequal. 

Every  proportion  is  thus  seen  to  be  an  equality  of  the  form 
a/b  =  c/d,  where  a,  6,  c,  and  d  are  certain  numbers.  These 
four  numbers  are  called  the  terms  of  the  proportion.  The 
first  and  fourth  (that  is,  a  and  d)  are  called  the  extremes 
of  the  proportion,  while  the  second  and  third  (b  and  c)  are 
called  the  means. 

Besides  writing  a  proportion  in  the  form  a/6  =  c/d,  it  may 
be  written  in  the  form  a  :  b  : :  c  :  d,  or  also  in  the  form  a :  d  = 
c:d.  In  all  cases  it  is  read  "  a  is  to  b  as  c  is  to  d,"  and  it 
means  that  the  fraction  a/6  equals  the  fraction  c/d. 


168  SECOND   COURSE   IN  ALGEBRA       [XVI,  §  96 

EXERCISES 

1.  Using  the  language  of  proportion,  read   each  of  the 
following  statements. 

(a)  i  =  f.  (c)   2: -1=8: -4. 

(6)  1:4  =  3:12.  (d)  £:£::4:3. 

2.  State  which  are.  the  extremes  and  which  the  means 
in  each  part  of  Ex.  1. 

3.  State  such  proportions  as  you  can  make  out  of  the 
following  four  quantities :    3  inches,  6  inches,  12  inches,  24 
inches. 

[HINT.    3  inches  is  to  6  inches  as  •  •  • .    Make  other  proportions  also.] 

4.  State  such  proportions  as  you  can  make  out  of  the 
following  four  quantities :  1  inch,  3  inches,  1  foot,  1  yard. 

[HINT.     First  express  all  quantities  in  inches.] 

5.  State  such  proportions  as  you  can  make  out  of  the 
following :   1  pint,  1  quart,  1  gallon,  2  gallons. 

6.  Do  as  in  Ex.  5  for  the  following :  2  seconds,  1  minute, 
1  hour,  a  day  and  a  quarter. 

7.  Do  as  in  Ex.  5  for  the  following :    1  cent,  1  dollar,  1 
centimeter,  1  meter. 

[HINT.     Compare  money  ratio  with  distance  ratio.] 

8.  Do  as  in  Ex.  5  for  the  following :    4  ounces,  1  pound, 
1  gallon,  1  quart. 

97.  Algebraic  Proportions.  If  we  consider  the  algebraic 
fraction  (a26)/(a&2),  we  see  (upon  dividing  both  numerator 
and  denominator  by  ab)  that  it  reduces  to  a/6.  In  other 
words,  we  have 

o26  =  a 
ab2    b 

This  is  an  example  of  an  algebraic  proportion.    Similarly, 
2x2y_  x 
4  xyz    2  z 


XVI,  §  98]  RATIO   AND   PROPORTION  169 

is  an  algebraic  proportion  and  may  be  written  if  desired  in 

the  form 

2  x2y  :  4  xyz  =  x  :  2  z. 

Likewise,  since 

a2-62 


__  _ 

we  have 


98.  Fundamental  Theorem.  Let  a/b  =  c/d  be  any  pro- 
portion. By  multiplying  both  sides  of  this  equality  by  bd, 
we  obtain 


or 

ad  =  be. 

This  result  may  be  stated  in  the  following  theorem. 

THEOREM  A.  In  any  proportion,  the  product  of  the  means 
is  equal  to  the  product  of  the  extremes. 

This  theorem  is  useful  in  testing  the  correctness  of  a 
proportion.  Thus  6:9=14:21  is  a  correct  proportion  be- 
cause the  product  of  the  means,  which  is  9  X  14,  is  equal  to 
the  product  of  the  extremes,  which  is  6  X21  ;  but  6  :  9  =  8  :  15 
is  not  correct  because  9  X8  is  not  equal  to  6  Xl5.  Similarly, 
x3  :  x2y  =  x:y,  because  x2y  -  x  =  xz  •  y. 

EXERCISES 

By  means  of  the  theorem  of  §  98,  test  the  correctness  of 
the  following  proportions. 

1.  5:6  =  15:18.  5.    17:19  =  21:23. 

2.  3:2  =  5:6.  6.   2  a  :  ab  =  W  x  :  5  bx. 

3.  4:  -1  =  8:  -2.        7.   3  m2  :  (a-b)  =6  m  :  2  m(a-b). 

4.  i:|  =  8:4.  8. 

9.     a2-62: 


170  SECOND    COURSE    IN   ALGEBRA       [XVI,  §  98 

By  means  of  the  theorem  of  §  98,  determine  the  value 
which  x  must  have  in  the  following  proportions. 

10.  z:4  =3:2. 

[HINT.     The  theorem  gives  4  •  3  =  x  •  2,  or  12  =  2  x.] 

11.  10:z  =  2:5.  13.    (z-5) :  4: :  2:  3. 

12.  25/32  =  8/z.  14.    (x-3)/(x-4)  =  5/6. 

15.  What  number  bears  the  same  ratio  to  4  as  16  does  to 
6? 

[HINT.  Let  x  represent  the  unknown  number  and  form  a  pro- 
portion.] 

16.  Divide  35  into  two  parts  whose  ratio  shall  be  f . 
[HINT.     Let  x  be  one  part.     Then  35  —  x  will  be  the  other  part.] 

17.  Divide  35  into  two  parts  such  that  the  lesser  dimin- 
ished by  4  is  to  the  larger  increased  by  9  as  1 :  3. 

18.  A  man's  income  from  two  investments  is  $980.     The 
two  investments  bear  interest  rates  which  are  in  the  ratio 
of  5  to  6.     What  income  does  he  receive  from  each? 

19.  Concrete  for  sidewalks  is  a  mixture  made  of  two  parts 
sand  to  one  part  cement.     How  much  of  each  is  required  to 
make  a  walk  containing  1500  cubic  feet? 

20.  Prove  that  no  four  consecutive  numbers,  as  n,  ft+1, 
ft +2,  ft +3,  can  form  a  proportion  in  the  order  given. 

99.  Application  to  Similar  Figures.  When  two  geometric 
figures  have  the  same  shape,  though  not  necessarily  the  same 
size,  they  are  called  similar  figures.  Thus  any  two  circles 
are  similar  figures;  likewise,  any  two  squares,  or  any  two 
cubes,  or  any  two  spheres. 

Two  triangles  may  be  similar,  as 
,  illustrated  in  Fig.  63. 

The  following  facts  are  shown   in 
/  ,    geometry  regarding  any  two  similar 
figures. 


XVI,  §99]  RATIO   AND   PROPORTION  171 

(a)  Corresponding  lines  are  proportional. 

Thus,  in  the  two  similar  triangles  of  Fig.  63,  if  the  side  AB  of  the 
one  is  twice  as  long  as  the  corresponding  side  A'B'  of  the  other,  then 
BC  is  twice  as  long  as  B'C'.  That  is, 

AB  =  BC 
A'B'     B'C'' 

In  the  same  way,  we  have  also 

AB       CA 


A'B'     C'A' 

(6)  Areas  are  proportional  to  the  squares  of  corresponding 
lines. 

Thus,  if  one  circle  has  a  radius  of  length  R  and  another  circle  has 
a  radius  of  length  r,  the  area,  A,  of  the  first  is  to  the  area,  a,  of  the 
second  as  R2  is  to  r2.  That  is  we  have  the  proportion  A  /a  =  R2/r2. 

(c)  Volumes  are  proportional  to  the  cubes  of  corresponding 
lines. 

Thus,  if  one  sphere  has  the  radius  R  and  another  has  the  radius  r, 
the  volumes  V  and  v  of  the  spheres  are  such  that  V/v  =  R3/r3. 

EXERCISES    ON    SIMILAR   FIGURES 

1.  In  the  two  similar  triangles  shown  in   §  99  suppose 
AB  =  2  feet,  A'B'  =  l  foot  4  inches,  and  BC  =  3  feet.     How 
long  will  B'C'  be? 

2.  If  a  tree  casts  a  shadow  40  feet  long  when  a  post  3£ 
feet  high  casts  a  shadow  4  feet  long,  how  high  is  the  tree  ? 

3.  Compare  the  areas  of  two  city  lots  of  the  same  shape 
if  a  side  of  the  one  is  three  times  as  long  as  the  corresponding 
side  of  the  other.     Does  it  matter  what  the  shape  of  each  is  ? 

4.  If  a  certain  bottle  holds  -J  pint,  how  much  will  a  bottle 
of  the  same  shape  but  only  half  as  high  hold  ? 


172  SECOND   COURSE   IN  ALGEBRA     [XVI,  §  100 

100.  Mean  Proportional.     If  the  means  of  a  proportion 
are  equal,  either  is  called  the  mean  proportional  between  the 
extremes. 

Thus  2  is  the  mean  proportional  between  1  and  4  because 
•£•=•}.  Likewise,  2  x  is  the  mean  proportional  between  x2  and  4, 
because  x2/2x  =2  z/4. 

NOTE.  The  mean  proportional  between  a  and  6  is  always  equal 
to  Vo6,  for  we  must  have  a/x=x/b.  Hence,  clearing  of  fractions, 
x2  =  ofc/andjtheref  ore  x  =  Vo6.  This  will  be  a  surd  (§  42)  unless  the 
product  ab  is  a  perfect  square.  For  example,  the  mean  proportional 
between  2  and  3  is  the  surd  V2  •  3,  or  V6  =2.44949  (table). 

101.  Third  and  Fourth  Proportionals.    The  third  propor- 
tional to  two  numbers  a  and  b  is  that  number  x  such  that 
a:  b  =  b:x. 

Thus  the  third  proportional  to  2  and  3  is  the  value  of  x  in  the 
equation  ^  ^ 

o  =  ~-     Solving,  £=•§•  =  4  \.     Arts. 

O        X 

The  fourth  proportional  to  three  numbers  a,  b,  and  c  is  that 
number  x  such  that  a:b  =  c:x. 

Thus  the  fourth  proportional  to  2,  3,  and  4  is  the  value  of  x  in 
the  equation  2/3  =  4/x.  Solving,  x  =  6.  Ans. 

EXERCISES 

1.  Find  the  mean  proportional  between  8  and  18. 
[HINT.     Let*  be  the  desired  mean.   Then8/x  =x/lS.  Solveforz.] 
Find  the  mean  proportional  between  each  of  the  follow- 

ing pairs  of  numbers.     In  cases  where  the  answer  is  a  numer- 
ical surd,  use  the  table  to  find  its  approximate  value. 

2.  9  and  81.  6.   |  and  |f. 

3.  6  and  7.  7.   2£  and  3^. 

4.  5  and  20.  8.   2  x2y  and  32  xy2. 

a2-2a-3 


K    K      A  10  Q        -  ^ 

6.   5  and  19.  9.  -  and 

a+1  a—  3 


XVI,  §  101] 


RATIO   AND   PROPORTION 


173 


Find  the  third  proportional  to  each  of  the  following  pairs. 

10.  3  and  4.  12.   2|  and  3^.        14.   z2-9  and  z-3. 

11.  18  and  50.        13.   2  x  and  x.         15.   2  and  6. 

Find  the  fourth  proportional  to  each  of  the  following  sets. 

16.  3,  4,  and  5.  19.    V2,  \/6,  and  A/12. 

17.  5,  4,  and  2.  20.   3  a,  2  6,  and  c. 

18.  2,  3^,  4i.  21.   x,  y,  and  xy. 

NOTE.  In  Exs.  16-21,  the  numbers  must  be  placed  in  the  pro- 
portion in  the  order  in  which  they  are  given,  as  in  the  illustrative 
examples  of  §  101. 

22.  In  the  semicircle  ABC  suppose 
DE  drawn  perpendicular  to  AB.    Then 
(as  shown  in  geometry)  the  length  of 
DE    will    be    the    mean    proportional    A 
between  the  lengths  of  AE  and  EB.  Fl<*-  64. 

If  A  E  =  4  inches  and  EB  =  16  inches,  find  DE. 

23.  The  figure  shows  a  circle  and  a  point  P  outside  it 
from  which  are  drawn  two  lines  PS  and  PT.    The  first  of 

p  these  lines  (called  a  secant)  cuts  through  the 
circle  at  two  points  R  and  S  while  the 
second  line  (called  a  tangent)  just  touches 
the  circle  at  the  point  T.  In  all  such  cases, 
the  tangent  length,  PT,  is  the  mean  pro- 
portional between  the  whole  secant,  PS,  and 
its  external  segment,  PR  (as  shown  in 
geometry). 
Find  the  length  of  PT  if  PR  =  4  and  RS=11. 

24.  If  a,  b,  c,  d  are  unequal  numbers  such  that  a:b  =  c:dt 
show  that  no  number  x  can  be  found  such  that 

a-\-x\  b+x  =  c-\-x  \d-\-x. 


FIG.  65. 


174  SECOND   COURSE    IN   ALGEBRA     [XVI,  §  102 

102.  Second  Fundamental  Theorem.     THEOREM  B.     // 
the  product  of  two  numbers  is  equal  to  the  product  of  two  other 
numbers,  either  pair  may  be  made  the  means  of  a  proportion 
in  which  the  other  two  are  taken  as  the  extremes. 

PROOF.  Suppose  mn  =  xy.  Dividing  both  members  by 
nx  gives  m/x  =  y/n,  or  m :  x  =  y  :  n,  which  is  one  of  the  possible 
proportions  mentioned  in  the  theorem. 

Similarly,  if  we  divide  both  members  of  mn  =  xy  by  ny  we 
obtain  m/y  =  x/n,  or  m:y  =  x:n,  which  is  another  of  the 
possible  proportions  mentioned  in  the  theorem. 

The  other  possible  proportions  are  x:m  =  n:y  and  n:x  = 
y :  m.  The  proof  of  these  is  left  to  the  pupil. 

For  example,  the  equality  2  •  9=3  •  6  gives  rise  to  the  propor- 
tions 2 :  3=6:  9,  2:6  =  3:9,  9:3=6:2,  and  9:  6=3:  2. 

103.  Inversion  in  a  Proportion.     THEOREM  C.     If  four 
quantities  are  in  proportion,  they  are  in  proportion  by  inversion; 
that  is  the  second  term  is  to  the  first  as  the  fourth  is  to  the 
third. 

PROOF.     We  are  to  show  that  if  a/b  =  c/d,  then  b/a  =  d/c. 
Since  a/b  =  c/d,  we  have,  by  Theorem  A,  ad  =  bc. 
Therefore,  by  Theorem  B,  we  may  write  b/a  =  d/c. 
For  example,  f  =  f  gives  by  inversion  the  new  proportion  f-  =  f . 

104.  Alternation  in  a  Proportion.     THEOREM  D.     //  four 
quantities  are  in  proportion,  they  are  in  proportion  by  alterna- 
tion; that  is  the  first  term  is  to  the  third  as  the  second  is  to 
the  fourth. 

The  proof  is  left  to  the  pupil.  First  use  Theorem  A.  See 
proof  of  Theorem  C. 

For  example,  f  =  f  gives  by  alternation  the  new  proportion  £  =f  • 

105.  Composition    in    a    Proportion.     THEOREM  E.     // 
four  quantities  are  in  proportion,  they  are  in  proportion  by 


XVI,  §  107]  RATIO   AND   PROPORTION  175 

composition;  that  is  the  sum  of  the  first  two  terms  is  to  the 
second  term  as  the  sum  of  the  last  two  terms  is  to  the  last 
term. 

PROOF.  We  are  to  show  that  if  a/b  =  c/d,  then  (a+b)/b  = 
(c+d)/d. 

Since  a/b  =  c/d,  we  may  add  1  to  each  member  of  this  equa- 
tion, thus  giving 

«+!=<  +  !,  which  reduces  to 


, 
b          a  o          a 

For  example,  f  =  f   gives   by  composition  the   new  proportion 

(2+3)/3  =  (6+9)/9,  orf=V5. 

106.  Division  in  a  Proportion.     THEOREM  F.     //  four 
quantities  are  in  proportion,  they  are  in  proportion  by  division; 
that  is,  the  difference  between  the  first  two  terms  is  to  the 
second  term  as  the  difference  between  the  last  two  terms  is  to 
the  last  term. 

PROOF.  We  are  to  show  that  if  a/b  =  c/d,  then  (a  —  &)/&  = 
(c-d)/d. 

Since  a/b  =  c/d,  -we  may  subtract  1  from  each  side  of  this 
equation,  thus  obtaining 

f-  1=^-1,  which  reduces  to  *LZ&=^. 
b          a  o          d 

For    example,    f  =f    gives    by    division    the    new    proportion 

(3-2)/2  =  (9-6)/6,  or|=f. 

107.  Composition  and  Division.     THEOREM  G.     //  four 
quantities  are  in  proportion,  they  are  in  proportion  by  composi- 
tion and  division;  that  is  the  sum  of  the  first  two  terms  is  to 
their  difference  as  the  sum  of  the  last  two  terms  is  to  their 
difference. 

PROOF.     We  are  to  show  that  if  a/b  =  c/d,  then 


a—b    c—d 


176  SECOND   COURSE   IN  ALGEBRA     [XVI,  §  107 

By  Theorems  E  and  F,  we  have 

a-j-b_c-\-d         ,    a—b_c  —  d 

~l       d~'  !       ~b       5" 

By  dividing  the  first  of  these  equations  by  the  second, 
member  by  member,  we  obtain  the  desired  result,  namely 


a—  b    c—d 

For  example,  f  =  |-  gives  by  composition  and  division  the-new  pro- 
portion (3  +2)  /(3-2)  =  (9  +6)  /(9-6),  or  f  =-^. 

108.  Several  Equal  Ratios.  THEOREM  H.  In  a  series  of 
equal  ratios,  the  sum  of  the  antecedents  is  to  the  sum  of  the  conse- 
quents as  any  antecedent  is  to  its  consequent. 

PROOF.  We  are  to  show  that  if  a/b  =  c/d  =  e/f=g/h=  •••, 
then 

a+c+e+g-\  —  _a_^c  _e_g  _ 
b+d+f+h+~-    b    d    f    h 
Let  k  be  the  value  of  any  one  of  the  equal  ratios,  so  that 

JL_«  £_c  r._e  k_g 
6'*    d'*   /     -A' 

=  fc&,  c  =  kd,  e  =  kf,  g  =  kh,  •••. 


Hence 


b+d+f+h+- 
or 

a+c+e+g-\  —  _^a_c  '  _e_Q  [_ 
b+d+f+h+-  b  d  f  h 

For  example,  the  three  equal  ratios  f  =  f  =  f  give  the  new  pro- 

portions 

2+4+6^2^.4,6 

3+6+9    369* 
or 

12  =  2=4=6 

18     3     6     9* 


XVI,  §  108]  RATIO   AND   PROPORTION  177 

EXERCISES 

1.  Given  the  proportion  f =J^-.    Write  down  the  vari- 
ous proportions  to  be  obtained  from  this  by  (1)  inversion, 
(2)  alternation,  (3)  composition,  (4)  division,  (5)  composi- 
tion and  division.     Note  that  each  new  proportion  thus  ob- 
tained is  a  true  one. 

2.  Show   that  if  a/b  =  c/d,  then   a2/b2  =  c2/d2',    in  other 
words,  if  four  numbers  are  in  proportion,  their  squares  are 
also  in  proportion. 

3.  Show  that  if  four  numbers  are  in  proportion,  their 
cubes  are  also  in  proportion;   likewise  their  square  roots; 
likewise  their  cube  roots. 

4.  Show  by  means  of  Theorem  A  (§  98)  that  a/b  =  a2/b2 
is  not  a  true  proportion,  unless  a  =  b.     In  other  words,  the 
ratio  of  two  numbers  is  not  in  general  the  same  as  the  ratio 
of  their  squares.     Prove  similarly  that  the  ratio  of  two  num- 
bers is  not  in  general  the  same  as  the  ratio  of  their  cubes, 
or  of  their  square  roots,  or  of  their  cube  roots. 

5.  If  a :  b  =  c  :  d,  establish  the  following  proportions. 

(a)  a2:b2c2  =  l:d2. 

SOLUTION.  It  suffices  to  show  here  that  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means,  for  if  these  two 
products  are  the  same,  the  proportion  in  question  is  true  by  Theorem 
B.  Thus  we  are  to  prove  that  azdz  =  62c2. 

Now,  we  know  (by  hypothesis)  that  a:b=c:d,  or  ad  =  bc. 
Squaring  gives,  as  desired,  a?d2  =  bzc2,  thus  completing  the  proof. 

(b)  ac:bd  =  c2:d2. 

[HINT.   Remember  to  use  the  hypothesis,  namely  that  a  :  b  =  c  :  d.] 

(c)  Vad:Vb  =  Vc:l.        (d)  a:a+b  =  a+c:  a+b+c+d.. 
(e)  a+b:  c+d=Va2+b2:  Vc2+d2. 

(/)  a+b+c+d:  a—b+c—d=a+b—c—d:  a—b  —  c+d. 
(g)  If  a  :  b  =  c  :  d,  and  x:y  =  z:w,  show  that  ax  :  cz  =  by  :  dw. 


CHAPTER  XVII 
VARIATION 

109.  Direct  Variation.  One  quantity  is  said  to  vary 
directly  as  another  when  the  two  are  so  related  that,  though 
the  quantities  themselves  may  change,  their  ratio  never 
changes. 

Thus  the  amount  of  work  a  man  does  varies  directly  as  the 
number  of  hours  he  works.  For  example,  if  it  takes  him  4  hours  to 
draw  10  loads  of  sand,  we  can  say  it  will  take  him  8  hours  to  draw  20 
loads.  Here  the  first  ratio  is  ^  and  the  second  is  -£$  and  the  two 
are  seen  to  be  equal,  though  the  numbers  in  the  second  have  been 
changed  from  what  they  were  in  the  first.  In  general,  if  the  man 
works  twice  as  long,  he  will  draw  twice  as  much ;  if  he  works  three 
times  as  long,  he  will  draw  three  times  as  much,  etc. ;  all  of  which 
implies  that  the  ratio  of  the  time  he  works  to  the  amount  he  draws 
in  that  time  never  changes. 

EXERCISES 

Determine  which  of  the  following  statements  are  true  and 
which  false,  giving  your  reason  in  each  instance. 

1.  The   amount  of  electricity  used  in   lighting   a  room 
varies  directly  as  the  number  of  lights  turned  on. 

2.  The   amount  of   water  in   a    cylindrical   pail  varies 
directly  as  the  height  to  which  the  water  stands  in  the 

pail. 

178 


XVII,  §  110]  VARIATION  179 

3.  The  amount  of  gasoline  used  by  an  automobile  in  any 
given  time  (one  week,  say)  varies  directly  as  the  amount 
of  driving  done. 

4.  The  time  it  takes  to  walk  from  one  place  to  another 
at  any  given  rate  (3  miles  an  hour,  say)  varies  directly  as  the 
distance  between  the  two  places. 

5.  The  time  it  takes  to  walk  any  given  distance  (5  miles, 
say)  varies  directly  as  the  rate  of  walking. 

6.  The  perimeter  of  a  square  varies  directly  as  the  length 
of  one  side. 

7.  The  circumference  of  a  circle  varies  directly  as  the 
length  of  the  radius. 

8.  The  area  of  a  square  varies  directly  as  the  length  of 
one  side. 

9.  x  varies  directly  as  10  x. 
10.   x  varies  directly  as  10  x2. 

110.  Inverse  Variation  One  quantity,  or  number,  is  said 
to  vary  inversely  as  another  when  the  two  are  so  related  that, 
though  the  quantities  themselves  may  change,  their  product 
never  changes. 

Thus  the  time  occupied  in  doing  any  given  piece  of  work  varies 
inversely  as  the  number  of  men  employed  to  do  it.  For  example, 
if  it  takes  2  men  6  days,  it  will  take  4  men  only  3  days.  The  point 
to  be  observed  here  is  that  the  first  product,  2X6,  equals  the  second 
product,  4X3.  In  general,  if  twice  as  many  men  are  employed  it 
will  take  half  as  long ;  if  three  times  as  many  men  are  employed,  it 
will  take  one  third  as  long,  etc.  In  all  these  cases,  the  number  of 
men  employed  multiplied  by  the  corresponding  time  required  to  do 
the  work  remains  the  same. 

NOTE.  The  term  varies  inversely  as  is  due  to  the  fact  that  in 
case  xy  never  changes  (as  required  by  the  above  definition),  it 
follows  that  x  +  (\/y)  never  changes,  since  xy=x  +  (\/y).  That 
is,  x  varies  directly  as  the  reciprocal,  or  inverse,  of  y  (§  109). 


180  SECOND   COURSE   IN   ALGEBRA          [XVII,  §  110 

EXERCISES 

Determine  which  of  the  following  statements  are  true  and 
which  false,  giving  your  reason  in  each  instance. 

1.  The  time  it  takes  water  to  drain  off  a  roof  varies  in- 
versely as  the  number  of  (equal  sized)  conductor  pipes. 

2.  The  time  it  takes  to  walk  any  given  distance  (5  miles, 
say)   varies  inversely  as  the  rate  of  walking.     (Compare 
Ex.  5,  p.  179.) 

3.  The  weight  of  a  pail  of  water  varies  inversely  as  the 
amount  of  water  that  has  been  poured  out  of  it. 

4.  x  varies  inversely  as  10/z. 
6.   x  varies  inversely  as  10/z2. 

111.  Joint  Variation.  One  quantity,  or  number,  is  said  to 
vary  jointly  as  two  others  when  it  varies  directly  as  their 
product. 

Thus  the  area  of  a  triangle  varies  jointly  as  its  base  and  altitude, 
for  if  A  be  the  area  of  any  triangle  and  b  its  base  and  h  its  altitude, 
we  have  A=^bh,  which  may  be  written  A/bh=%.  Whence  A 
varies  directly  as  the  product  bh  (§  109),  that  is  the  ratio  of  A  to 
bh  is  always  the  same,  namely  ^  in  this  instance. 

EXERCISES 

I 

Determine  whether  the  following  statements  are  true, 
giving  your  reason  in  each  instance. 

1.  The   area  of   a  rectangle   varies   jointly   as  its  two 
dimensions,  that  is  as  its  length  and  breadth. 

2.  The  pay  received  by  a  workman  varies  jointly  as  his 
daily  wage  and  the  number  of  days  he  works. 

3.  The  amount  of  reading  matter  in  a  book  varies  jointly 
as  the  thickness  of  the  book  and  the  distance  between  the 
lines  of  print  on  the  page. 


XVII,  §  113]  VARIATION  181 

4.  The  interest  received  in  one  year  from  an  investment 
varies  jointly  as  the  principal  and  rate. 

5.  The  volume  of  a'  rectangular  parallelepiped   (such  as 
an  ordinary  rectangular   shaped  box)  varies   jointly  as  its 
length,  breadth,  and  height. 

[HINT.  Here  we  have  one  quantity  varying  jointly  as  three 
others.  First  make  a  definition  yourself  of  what  such  variation 
means.] 

112.  Variables  and  Constants.  When  we  say  that  the 
amount  of  work  a  man  does  varies  directly  as  the  number  of 
hours  he  works  (see  §  109),  we  are  dealing  with  two  quanti- 
ties, namely  the  amount  of  work  done  and  the  time  used  in 
doing  it.  But  it  is  to  be  observed  that  these  are  not  being 
regarded  as  fixed  quantities,  but  rather  as  changeable  ones, 
the  only  essential  idea  being  that  their  ratio  never  changes. 
In  general,  quantities  which  are  thus  changeable  throughout 
any  discussion  or  problem  are  called  variables,  while  quan- 
tities which  do  not  change  are  called  constants. 

113.  The  Different  Types  of  Variation  Stated  as  Equa- 
tions. We  may  now  state  very  briefly  and  concisely  what  is 
meant  by  the  different  types  of  variation  mentioned  in 
§§  109-111  and  certain  other  important  types  also.  To  do 
this,  let  us  think  of  x,  y,  and  z  as  being  certain  variables 
and  k  as  being  some  constant.  Then 

(1)  To  say  that  x  varies  directly  as  y  means  (by  §  109)  that 

T 

-  =  fc,     or    x  =  ky. 

y 

(2)  To  say  that  x  varies  inversely  as  y  means  (by  §  110)  that 

xy  —  k,     or    x  =  -. 

y 

(3)  To  say  that  x  varies  jointly  as  y  and  z  means  (by  §  1 11)  that 

1* 

—  =  k,    or    x  —  kyz. 
yz 


182  SECOND    COURSE    IN   ALGEBRA     [XVII,  §  113 

Two  other  important  types  of  variation  are  described 
below : 

(4)  To  say  that  x  varies  directly  as  the  square  of  y  means  that 

-  =  k,     or     x  =  ky2. 

(5)  To  say  that  x  varies  inversely  as  the  square  of  y  means  that 

xy2  =  k,     or    x  =  —  • 

y2 

In  all  these  types  of  variation  it  is  important  to  observe 
that  the  value  which  must  be  given  to  the  constant  k  depends 
upon  the  particular  statement  or  problem  in  hand.  For 
example,  consider  the  statement  that  "  The  area  of  a  rec- 
tangle varies  jointly  as  its  two  dimensions."  This  means 
(see  (3))  that  if  we  let  A  be  the  variable  area  and  a  and  b 
the  variable  dimensions,  then  A  =  kab.  But  in  this  case  we 
know  by  arithmetic  that  A  =  ab,  so  the  value  of  k  here  must 
be  1.  On  the  other  hand,  consider  the  statement  that  "  The 
area  of  a  triangle  varies  jointly  as  its  base  and  altitude." 
Letting  A  be  the  variable  area  and  b  and  h  the  variable  base 
and  altitude,  respectively,  this  means  that  A  =  kbh.  But 
here,  as  we  know  from  arithmetic,  k  =  \. 

EXERCISES 

Convert  each  of  the  following  statements  into  equations, 
supplying  for  each  the  proper  value  for  the  constant  k 
mentioned  in  §  113. 

1.  The  circumference  of  a  circle  varies  directly  as  the 
radius. 

[HINT.     Let  C  stand  for  circumference  and  r  for  radius.] 

2.  The   circumference  of  a  circle  varies  directly  as  the 
diameter. 

3.  The  area  of  a  circle  varies  directly  as  the  square  of  the 
radius. 


XVIT,  §  113]  VARIATION  183 

4.  The  area  of  a  circle  varies  directly  as  the  square  of  the 
diameter. 

5.  The  area  of  a  sphere  varies  directly  as  the  square  of 
the  radius.     (See  §  14,  (/).) 

6.  The  volume  of  a  rectangular  parallelepiped  varies 
jointly  as  its  length,  breadth,  and  height.     (See  Ex.  5,  p.  181.) 

7.  Interest  varies  jointly  as  the  principal,  rate,  and  time. 

8.  The  volume  of  a  sphere  varies  directly  as  the  cube 
of  the  radius. 

[HINT.  First  supply  for  yourself  the  definition  of  what  this 
type  of  variation  means.] 

9.  The  volume  of  a  circular  cone  varies  jointly  as  the  alti- 
tude and  the  square  of  the  radius  of  the  base.     (See  p.  103.) 

10.  The  distance,  measured  in  feet,  through  which  a  body 
falls  if  dropped  vertically  downward  from  a  position  of  rest 
(as  from  a  window  ledge)  varies  directly  as  the  square  of  the 
number  of  seconds  it  has  been  falling. 

[HINT.  It  is  found  by  experiments  in  physics  that  the  value  of 
the  constant  A;  is  in  this  case  32  (approximately).] 

11.  The  following,  like  Ex.   10,  are  statements  of  well- 
known  physical  laws.     Convert  each  into  an  equation  with- 
out, however,  attempting  to  supply  the  proper  value  of  k, 
since  to  do  so  requires  a  study  of  physics. 

(a)  If  a  body  is  tied  to  a  string  and  swung  round  and  round 
in  a  circle  (as  in  swinging  a  pail  of  water  at  arm's  length 
from  the  shoulder) ,  the  force,  F,  with  which  it  pulls  outward 
from  the  center  (called  centrifugal  force)  varies  directly  as  the 
square  of  the  velocity  of  the  motion. 

(6)  The  intensity  of  the  illumination  due  to  any  small 
source  of  light  (such  as  a  candle)  varies  inversely  as  the  square 
of  the  distance  of  the  object  illuminated  from  the  source  of 
light. 


184 


SECOND   COURSE    IN  ALGEBRA      [XVII,  §  113 


(c)  When  an  elastic  string  is  stretched  out,  as  represented 
in  Fig.  66,  the  tension  (force  tending  to  pull  it  apart  at  any 
point)  varies  directly  as  the  length  to  which  the  string  has 
been  stretched  (Hooke's  Law). 


FIG.  66. 

(d)  The  pressure  per  square  inch  which  a  given  amount 
of  gas  (such  as  air,  or  hydrogen,  or  oxygen,  or  illuminating 
gas)  exerts  upon  the  sides  of  the  receptacle  which  holds 

the  gas  (such  as  a  bag)  varies 
inversely  as  the  volume  of  the 
receptacle  (Boyle's  Law). 


FIG.  67. 


For  example,  whenever  air  is  con- 
fined in  a  rubber  balloon,  as  in  the  first 
drawing  in  Fig.  67,  it  exerts  a  certain 
pressure  upon  each  square  inch  of  the 
interior  surface.  If  the  balloon  be 
squeezed,  as  in  the  second  drawing 
in  Fig.  67  (no  air  being  allowed  to  escape),  until  its  volume  is  half 
of  what  it  was  before,  this  pressure  will  be  exactly  doubled. 

114.  Problems  in  Variation.  The  problems  naturally 
arising  in  the  study  of  variation  fall  into  two  general  classes 
as  follows : 

(1)  Those  in  which  the  value  of  the  constant  k  mentioned 
in  §  113  can  be  determined  from  the  statement  of  the  problem 
and  forms  an  essential  part  in  the  solution.  This  kind  of 
problem  is  illustrated  by  Exs.  1-10  on  pages  185-187.  The 
solution  given  for  Ex.  1  should  be  well  understood  before  the 
pupil  undertakes  Exs.  2-10. 


XVII,  §  114]  VARIATION  185 

(2)  Those  in  which  it  is  not  necessary  to  know  the  value  of 
k.  Such  problems  are  illustrated  in  Exs.  11-20,  pp.  187-189. 

The  pupil  is  advised  to  work  several  problems  from  each 
group  rather  than  to  confine  his  attention  to  either. 

EXERCISES 

I.  ILLUSTRATIONS  OF  CASE  (1) 

1.  In  a  fleet  of  ships  all  made  from  the  same  model  (that 
is,  of  the  same  shape,  but  of  different  sizes)  the  area  of  the 
deck  varies  directly  as  the  square  of  the  length  of  the  ship. 
If  the  ship  whose  length  is  200  feet  has  5000  square  feet  of 
deck,  how  many  square  feet  in  the  deck  of  the  ship  which 
is  300  feet  long? 

SOLUTION.  Let  A  represent  the  area  of  deck  on  the  ship  whose 
length  is  I.  Then  the  given  law  of  variation,  expressed  as  an 
equation  (§  113),  is 

(1)  A=klz.     (k  =  some  constant) 

Since  the  ship  which  is  200  feet  long  has  5000  square  feet  of  deck, 
it  follows  from  (1)  that  we  must  have 

=  /c(200)2. 


This  equation  tells  us  that  the  value  of  k  in  the  present  problem 
must  be 

.5000  =     5000      =1 
(200)2     200X200     8* 

Placing  this  value  of  k  in  (1),  gives  us  an  equation  which  deter- 
mines completely  the  relation  between  A  and  I  in  the  present  problem, 
that  is 
(2)  A  =  iZ*. 

Now  the  problem  asks  how  many  square  feet  of  deck  there  are 
in  the  ship  whose  length  is  300  feet.  This  can  be  found  by  simply 
placing  £  =  300  in  (2)  and  solving  for  A.  Thus 

A  =4  X(30Q)2=300*3QP:=  11,250  square  feet.     Ans. 

O  O 


186  SECOND   COURSE    IN   ALGEBRA     [XVII,  §  114 

NOTE.  Observe  that  the  first  step  in  the  above  solution  is  to 
express  as  an  equation  the  law  of  variation  belonging  to  the  problem. 
Next,  the  constant  k  is  determined.  After  this,  the  first  equation 
is  rewritten  in  its  more  exact  form  obtained  by  assigning  to  k  its 
value.  The  answer  is  then  readily  obtained. 

These  steps  should  be  followed  in  working  each  of  the  Exs.  2-10. 

2.  In  a  fleet  of  ships  all  of  the  same  model,  the  ship  whose 
length  is  200  feet  contains  6000  square  feet  in  its  deck.   How 
long  must  a  similar  ship  be  made  if  its  deck  is  to  contain 
13,500  square  feet? 

3.  To  make  a  suit  of  clothes  for  a  man  who  is  5  feet 
8  inches  high  requires  6  square  yards  of  cloth.     How  much 
cloth  will  be  required  to  make  a  suit  for  a  man  of  similar  build, 
whose  height  is  6  feet  2  inches? 

[HINT.  The  areas  of  any  two  similar  figures  vary  directly  as 
the  squares  of  their  heights.] 

4.  If  10  men  can  do  a  piece  of  work  in  20  days,  how  long 
will  it  take  25  men  to  do  it? 

[HINT.  The  time  required  varies  inversely  as  the  number  of 
men  employed.] 

5.  The  horsepower  required  to  propel  a  ship  varies  di- 
rectly as  the  cube  of  the  speed.     If  the  horsepower  is  2000 
at  a  speed  of  10  knots,  what  will  it  be  at  a  speed  of  15 
knots? 

6.  A  silver  loving-cup  (such  as  is  sometimes  given  as  a 
prize  in  athletic  contests)  is  to  be  made,  and  a  model  is  first 
prepared  out  of  wood.     The  model  is  8  inches  high  and 
weighs  12  ounces.     What  will  the  loving-cup  cost  if  made 
10  inches  high,  it  being  given  that  silver  is  17  times  as  heavy 
as  wood  and  costs  $2.20  an  ounce? 

[HINT.  The  volumes  and  hence  the  weights  of  any  two  similar 
figures  vary  directly  as  the  cubes  of  their  heights.  See  §  99  (c).] 


XVII,  §  114]  VARIATION  187 

7.  When  electricity  flows  through  a  wire,  the  wire  offers 
a  certain  resistance  to  its  passage.     The  unit  of  this  resist- 
ance is  called  the  ohm,  and  for  a  given  length  of  wire  the 
resistance  varies  inversely  as  the  square  of  the  diameter.     If 
.a  certain  length  of  wire  whose  diameter  is  \  inch  offers  a 
resistance  of  3  ohms,  what  will  be  the  resistance  of  a  similar 
wire  (same  length  and  material)  ^  of  an  inch  in  diameter? 

8.  Three  spheres  of  lead  whose  radii  are  6  inches,  8  inches, 
and  10  inches  respectively  are  melted  and  made  into  one. 
What  is  the  radius  of  the  resulting  sphere  ? 

9.  On  board  a  ship  at  sea  the  distance  of  the  horizon 
varies  directly  as  the  square  root  of  one's  height  above  the 
water.     If,  at  a  height  of  20  feet,  the  horizon  is  5.5  miles  dis- 
tant, what  is  its  distance  as  seen  from  a  light-house  80  feet 
above  sea-level? 

10.  The  horsepower  that   a   shaft   can   safely  transmit 
varies  jointly  as  its  speed  in  revolutions  per  minute  and  the 
cube  of  its  diameter.     A  3-inch  steel  shaft  making  100  revo- 
lutions per  minute  can  transmit  85  horsepower.     How  many 
horsepower  can  a  4-inch  shaft  transmit  at  a  speed  of  150 
revolutions  per  minute  ? 

II.  ILLUSTRATIONS  OF  CASE  (2) 

11.  Knowing  that  the  force  of  gravitation  due  to  the 
earth  varies  inversely  as  the  square  of  the  distance  from  the 
earth's  center  (Newton's  Law  of  Gravitation),  find  how  far 
above  the  earth's  surface  a  body  must  be  taken  in  order  to 
lose  half  its  weight. 

SOLUTION.  Letting  W  represent  the  weight  of  a  given  body  at  the 
distance  d  from  the  earth's  center,  the  law  stated  above,  when  ex- 
pressed as  an  equation,  becomes 

( 1 )  W  =  -  -     (k  =  some  constant) 

a* 


188  SECOND   COURSE   IN  ALGEBRA     [XVII,  §  114 

Now  let  Wi  represent  the  weight  of  the  body  when  on  the  surface. 
Remembering  that  the  earth's  radius  is  4000  miles  (approximately), 
equation  (1)  gives 

(2)  k 


40002 

Next,  let  x  represent  the  desired  distance,  namely  the  distance 
above  the  surface  at  which  the  same  body  loses  half  its  weight. 
At  this  distance  its  weight  will  consequently  be  %Wi,  while  its  dis- 
tance from  the  earth's  center  is  now  4000  +x.  So  (1)  gives 

(3)  IKi  = , & 

2       (4000  +z)2 

Dividing  equation  (3)  by  equation  (2),  noting  the  cancelation  of 
W\  on  the  left  and  of  the  (unknown)  k  on  the  right,  we  obtain 

1  =      400Q2 

2  (4000  +xY 

It  remains  only  to  solve  this  equation  for  x. 
Clearing  of  fractions,  (4000  +z)2  =  2  -  40002  =  40002  •  2. 
Extracting  the  square  root  of  both  members,  4000  +x  =4000 \/2. 
Solving,         x  =4000 V2  -4000  =4000( V2 -1)  miles.     Ans. 
To  find  the  approximate  value  of  this  answer,  we  have  (see  table) 

V2  =  1.41421 
so  that  x  =4000(1.41421  -1)  =  4000  X. 41421  =  1656.84  miles.      Ans. 

NOTE.  Observe  that  the  first  step  in  the  above  solution  (as 
also  in  the  preceding  exercises)  is  to. express  as  an  equation  the  law 
of  variation  belonging  to  the  problem.  Then  write  down  the  two 
special  equations  which  express  the  particular  conditions  given  in 
the  problem  and  divide  one  of  these  equations  by  the  other  to 
eliminate  the  unknown  k.  The  answer  is  then  readily  obtained. 
A  similar  process  should  be  followed  in  working  the  remaining 
exercises  of  this  list. 

12.  Show  that  the  earth's  attraction  at  a  point  on  the  sur- 
face is  over  5000  times  as  strong  as  at  the  distance  of  the 
moon,  that  is  at  the  (approximate)  distance  of  280,000 
miles. 

[HINT.  Call  W\  the  weight  of  a  given  body  on  the  surface,  and 
let  W2  represent  the  weight  of  the  same  body  at  the  distance  of  the 
moon  from  the  earth's  center.  Then  use  the  law  expressed  in  (1) 
of  the  solution  of  Ex.  11.1 


XVII,  §  114]  VARIATION  189 

13.  A  book  is  being  held  at  a  distance  of  2  feet  from  an 
incandescent  lamp.     How  much  nearer  must  it  be  brought 
in  order  that  the  illumination  on  the  page  shall  be  doubled? 
(See  Ex.  11  (6),  p.  183.) 

14.  If  two  like  coins  (such  as  quarter  dollars)  were  melted 
and  made  into  a  single  coin  of  the  same  thickness  as  the  origi- 
nal, show  that  its  diameter  would  be  \/2  times  as  great. 

[HINT.  Call  D  the  diameter  of  the  given  coins  and  A  the  area 
of  each.  Note  that  the  area  of  the  new  coin  will  then  be  2  A.  Use 
the  result  stated  in  Ex.  3,  p.  182.] 

15.  Find  the  result  in  Ex.  14  when  four  equal-sized  coins 
are  used. 

16.  Show  that  a  falling  body  will  pass  over  the  second 
3  feet  of  its  descent  in  about  .4  of  the  time  it  takes  it  to 
pass  over  the  first  3  feet.     (See  Ex.  10,  p.  183.) 

17.  The  time  required  for  a  pendulum  to  make  a  complete 
oscillation  (swing  forward  and  back)  varies  directly  as  the 
square  root  of  its  length.     By  how  much  must  a  2-foot  pendu- 
lum be  shortened  in  order  that  its  time  of  complete  oscilla- 
tion may  be  halved  ? 

18.  If  the  diameter  of  a  sphere  be  increased  by  10%,  by 
what  per  cent  will  the  volume  be  increased  ? 

19.  Show  that  if  a  city  is  receiving  its  water  supply  by 
means  of  a  main  (large  pipe)   from  a  reservoir,  the  supply 
can  be  increased  25%  by  increasing  the  diameter  of  the  main 
by  about  12%. 

20.  It  is  desired  to  build  a  ship  similar  in  shape  to  one 
already  in  use  but  having  a  40%  greater  cargo  space  (or  hold). 
By  what  per  cent  must  the  beam  (width  of  the  ship)  be  in- 
creased.    (See  §  99  (c).) 


190 


SECOND    COURSE    IN   ALGEBRA     [XVII,  §  115 


115.  Variation  Geometrically  Considered.  If  a  variable 
y  varies  directly  as  another  variable  x,  we  know  (§  113)  that 
this  is  equivalent  to  having  the  equation  y  =  kx,  where  k  is 
some  constant.  If  the  value  of  &  is  1,  this  equation  takes  the 
definite  form  y  =  x,  and  we  may  now  draw  its  graph,  the 
result  being  a  certain  straight  line.  If,  on  the  other  hand, 
k  =  2,  we  have  y  =  2  x}  and  this  again  is  an  equation  whose 


234 
FIG.  68.  —  DIRECT  VARIATION. 


graph  may  be  drawn,  leading  to  a  straight  line,  but  a  differ- 
ent one.  In  general,  whatever  the  value  of  jfe,  the  corre- 
sponding equation  has  a  straight-line  graph.  The  fact  that  in 
all  cases  the  graph  is  a  straight  line  characterizes  this  type 
of  variation,  that  is,  characterizes  the  type  in  which  one 
variable  varies  directly  as  another.  The  figure  shows  the 
lines  corresponding  to  several  different  values  of  k. 


XVII,  §  115] 


VARIATION 


191 


In  case  a  variable  y  varies  inversely  as  another  variable  x, 
we  know  (§  113)  that  there  exists  an  equation  of  the  form 
y  =  k/x,  where  k  is  some  constant.  If  we  let  fc  =  l,  this  be- 
comes y=l/x.  By  letting  x  take  a  series  of  values  and 
determining  the  corresponding  values  of  y  from  this  equa- 
tion (thus  forming  a  table  as  in  §  57)  we  obtain  the  graph. 
Similarly,  corresponding  to  the  value  A;  =  2  we  have  y  =  2/x, 


1  2  3  4  X 

FIG.  69.  —  INVERSE  VARIATION. 

and  this  equation  has  a  definite  graph  which  is  different  from 
the  one  just  mentioned.  In  general,  whatever  the  value  of  fc, 
the  corresponding  equation  has  a  graph,  but  it  is  now  to  be 
noted  that  these  graphs  are  not  straight  lines;  they  are 
hyperbolas.  (See  Ex.  2,  §  78.)  The  figure  shows  the  curves 
corresponding  to  several  different  values  of  k. 

NOTE.     Though  these  curves  differ  in  form,  they  have  the  follow- 
ing feature  in  common :   Through  the  origin  draw  any  two  straight 


192  SECOND   COURSE    IN   ALGEBRA      [XVII,  §  115 

lines  (dotted  in  figure).  Then  the  intercepted  arcs  AB,  CD,  EF, 
GH,  etc.,  are  similar,  that  is  the  smallest  arc  when  simply  magnified 
by  the  proper  amount  produces  one  of  the  others. 

EXERCISES 

Draw  diagrams  to  represent  the  geometric  meaning  of 
each  of  the  following  statements. 

1.  y  varies  directly  as  the  square  of  x. 

2.  y  varies  inversely  as  the  square  of  x. 

3.  y  varies  as  the  cube  of  x. 

4.  y  varies  directly  as  x,  and  y  =  6  when  x  =  2. 
[HINT.     The  diagram  here  consists  of  a  single  line.] 

5.  y  varies  inversely  as  x,  and  y  =  6  when  x  =  2. 

6.  The  cost  of  n  pounds  of  butter  at  40j£  per  pound  is 
c  =  40  n. 

7.  The  amount  of  the  extension,  e,  of  a  stretched  string 
is  proportional  to  the  tension,  t,  and  e  =  2  in.  when  t  =  10  Ib. 
(See  Ex.  11  (c),  p.  184.) 

8.  The  pressure,  p,  of  a  gas  on  the  walls  of  a  retaining 
vessel  varies  inversely  as  the  volume,  v ;    and  p  =  40  Ib.  per 
square  foot  when  v  =  10  cu.  ft. 

9.  The  length,  L,  of  any  object  in  centimeters  is  propor- 
tional to  its  length,  I,  expressed  in  inches;     and  L  =  2.54 
when  1=1. 


NEWTON 
(Sir  Isaac  Newton,  1642-1727) 

Discoverer  of  the  law  of  gravitation  and  famous  in  algebra  for  his  discov- 
ery of  the  binomial  theorem.  Inventor  of  the  branch  of  higher  mathematics 
called  the  Calculus,  wherein  rates  of  motion  and  other  changing,  or  variable, 
quantities  are  extensively  studied. 


CHAPTER  XVIII 
EXPONENTS 

I.   POSITIVE  INTEGRAL  EXPONENTS 

116,  Powers.    Involution.    Just  as  a2 = u  -  a ;  a3  =  a  -a  •  a ; 
etc.,  so  we  define  the  nth  power  of  a,  where  n  is  any  posi- 
tive integer,  as  follows : 

an  =  a'Ci'a'a'a--a  (n  factors) . 

The  process  of  finding  the  power  of  a  number,  or  expression, 
is  called  involution. 

117.  Laws  of  Exponents.     There  are  five  fundamental 
laws  of  exponents  which  are  as  follows,  it  being  understood 
that  m  and  n  everywhere  stand  for  positive  integers : 

I.   MULTIPLICATION  LAW.     This  law  for  multiplying  two 
powers  of  the  same  quantity  is 

Qm  .  ^n=-^7n+n< 

PROOF. 

am  =  a°  a  -  a  •  a  •  a  •••  a  (m  factors).  (§  116) 

an  —  a-  a-  a-  a  •  a  •••  a  (n  factors). 
Hence 

am  .  an={a  •  a-  a  •  a  •••  a  (m  factors)}  •  {a  •  a  •  a  •  a  •••  a  (n  factors)} 
=  a-  a-  a  -  a---  a  (m-\-n)  f actors  =  am+n.  (§  116) 

Therefore 

am  -  an  =  am+n. 
ILLUSTRATIONS. 


194  SECOND   COURSE    IN   ALGEBRA     [XVIII,  §  117 

II.  DIVISION  LAW.     The  law  for  dividing  one  power  by 
another  power  of  the  same  quantity  is 

am  -^-  an  =  am~n  . 

PROOF.      am  -=-  a»  =  —  =  *  '  *  '  *  '  **'"  a  '  a  '  a'"  a  (m  factor3) 

an  tf  •  tf  •  fa  •  •  •  JL  (n  factors) 

=  a  •  a  •  a  •  a  •••  a  (m—ri)  factors  =  am~n.  (§  116) 
Therefore 

am  -j-on  =  am~n. 
ILLUSTRATIONS. 

36  ^.32  =36-2  =34  .        (  _2)5  +  (  _2)3  =  (  -2)2  ; 

x8  -r-z5  =x3  •  (a  +b)7  -7-  (a  +6)3  =  (a  +6)4. 

III.  LAW  FOR  THE  POWER  OF  A  POWER.     The  law  for 
raising  a  power  of  a  quantity  to  a  new  power  is 


mn 


a 
PROOF.         (am)n  =  am  •  am  •  am  •  am  -•  a™  (n  factors)  (§  116) 

=  am+m+m+m  +  "-   +m  (n  terms)^  (La\V   I) 

Therefore  (am}n  =  amn  since  m  +m  -\-m  +  •  •  •  +m  to  n  terms  =  ran. 
ILLUSTRATIONS. 

(42)3=42X3=46;       £(_2)3}3  =  (_2)9; 
(X4)5=:C20;  {(a  +6)3)4  =  (a  +6)12. 

IV.   LAW  FOR  THE  POWER  OF  A  PRODUCT.     The  law  for 
raising  to  a  power  a  product  of  two  quantities  is 

(ab)n  =  anbn. 
PROOF. 


(aZ>)  •  (afc)  •••  (06)  (n  factors)  (§116) 

{a  •  a  •  a  •••  a  (n  factors)}  •  [b  •  b  •  b  •  b  •••  b  (n  factors)} 
anbn. 


Therefore 
ILLUSTRATIONS. 


(2  X3)4  =  24  X34  ;       {(  -3)  (  -2)}3  =  (  -3)3(  -2)3  ; 
{(a  +6)  (c  +d)}3  =  (a  +6)3 


XVI11,  §  117]  EXPONENTS  195 

V.   LAW  FOR  THE  POWER  OF  A  QUOTIENT.     The  law  for 
raising  to  a  power  the  quotient  of  two  quantities  is 


Jb, 
PROOF.  Mn»/|j  .  M  .  /^  ...  ^\  (n  factors)  (§  116) 


a  (n  factors)     a 


b  -  b  -  b  •••  b  (n  factors)     bn 
Therefore 


ILLUSTRATIONS.  (|V~fJ;       f-^V 

\o/        o  \   o    /  o" 

(x\7_x^.        /a  -|-b\5  _  (q-j-b)5 
W       Z/7'       \a  —  b)       (a—  6)5 

EXERCISES 

Find  the  results  of  the  indicated  operations  in  the  following 
cases,  using  one  (or  more)  of  the  five  laws  in  §  117. 

1.  25  •  23.  8.  I2a  •  ta.  15.  xl°  +  x2. 

2(        1  "\3       /        1  "\2  Q      ?r— 1       /yr+1  1C      -»>jl2   •   ^yj6 

.     ^ — i)      •   \ — L)   .  o.    Z          •  Z        .  lu.    7/fr     "!~7fl  . 

4/y.lO    .7.2  -11       /7P+g— 1    .  /^l+r  1 Q      nm,^_rA 

.   ju        JL  ,  XJ..    ;/  y      .  xo.    ly      .  y  . 

5.  m12  -m13.  12.  83-^82.  19.  i2a^-^. 

6.  yb-yn.  13.  (-3)5^(-3)3.     20.  z^+z*-1. 

22.  0w+p-1-7-01+'\  25.   S(-8)3J4. 

23.  (a+b)2r+(a+b)r-1.      26.   (x6)4.  28.   (m4)8. 

24.  (25)3.  27.   (?/3)7. 

29.  (a26)3. 

[HINT  ro  Ex.  29.     First  use  Law  IV,  then  Law  III.] 

30.  (z3!/2)2.       32.   (a263c)4.          34.   S(a+6)2(c+d)3|4. 

31.  (a6c)3.        33.   (ra2n3w4)3.      35.   (xY)2m-      36.   (r2< 


o\3  /»>?5\4  /rn2 

37.  f  ?  )  -  38.     ^  )  .  39. 

n  J 


.  . 

V  n  J  \s 


196  SECOND    COURSE    IN   ALGEBRA        [XVIII,  §  117 

40.  f*-aY-     .  42.  c^y  Y^y.     44.  f-^v. 

V2/y      V2/V  \      «V 

43.  f-Y-^C-9Y-  45.  (-  — 

W      \y*J  V     2/3w 


118.  Roots.  Evolution.  Just  as  VcT means  the  number 
whose  square  gives  a,  and  Vo  means  the  number  whose  cube 
gives  a,  etc.,  so  we  define  the  nth  root  of  a,  Va,  to  be  the 
number  whose  nth  power  gives  a,  that  is  we  agree  that 


Thus  Vxl2=x4, because  (z4)3=z12.  (§  117,  Law  III.)  Similarly 
Va1668  =  a462,  because  (ct462)4  =  ct16b8. 

NOTE.     In  case  n  =  2,  we  write  simply  V"  instead  of  v/  . 

The  number  n  is  called  the  incfex  of  the  root. 
The  number  under  the  sign  V~,  as  a,  is  called  the  radicand. 
The  process  of  finding  the  root  of  a  number  or  expression 
is  called  evolution. 

119.  Rule  for  Finding  the  nth  Root  of  an  Expression.  The 
nth  root  of  an  expression  may  be  obtained  readily  in  case  the 
expression  itself  is  an  exact  nth  power.  This  is  illustrated 
in  the  following  examples. 

EXAMPLE  1.     To  find  the  value  of  \^m*n9. 

SOLUTION.  The  expression  men9  may  be  written  as  an  exact 
cube,  namely  (m2n3)3.  (Laws  IV  and  III  of  §  117.) 

Therefore   \/m6n9=  V(m2n3)3  =  m2w3.     Ans.  (§118) 

EXAMPLE  2.     To  find  the  value  of 

SOLUTION.     The  expression  x  V  may  be  written  as  an  exact  5th 

z15 

power,   namely  f  ^)  -     (Laws  IV  and  III  of  §  117.) 

Therefore   A/^-6==A/(^V=^-     Ans.  (§118) 


XVIII,  §  119] 


EXPONENTS 


197 


Observe  that  the  answer  to  Example  1  (namely  ra2n3)  is 
the  result  of  simply  dividing  each  exponent  of  the  radicand 
(namely  w6n9)  by  the  index  of  the  desired  root  (namely  3). 
Similarly,  in  Ex.  2  if  we  simply  divide  each  exponent  in 

f-  by  5  we  get  the  answer  immediately.     Thus,  in  practice, 

& 

we  use  the  following  rule. 

To  find  the  nth  root  of  an  exact  nth  power,  divide  the  expo- 
nent of  each  factor  of  the  radicand  by  n. 

Thus 


NOTE.  It  jwill  be  recalled  (§  39)  that,  unless  otherwise  stated, 
the  symbol  Va  means  the  positive  number  whose  square  is  a.  Thus 
V9=+3,  the  other  root,  —3,  being  represented  by  —  V9.  This 
agreement  is  made  in  order  to  bring  about  perfect  definiteness  in  the 
use  of  the  symbol  V  . 

EXERCISES 


Determine  (from  the  definition  in  §  1  18)  the  value  of  : 

_ 


a 
625. 


2.  V-27. 

3.  -v/SL  6.  \/A.  8. 
Determine  by  means  of  the  Rule  in  §  119  the  value  of: 

10.   ^64  a666.         [HINT.    Write  26  for  64.] 

23. 
24. 


11.  A/625  a864. 

12.  \/-27m?n«. 

13.  - 

14.  - 


15. 


22. 


198  SECOND    COURSE    IN   ALGEBRA      [XVIII,  §  120 

II.   FRACTIONAL,  ZERO,  AND  NEGATIVE  EXPONENTS 

120.  Introduction  of  General  Exponents.     Thus  far  we 
have    considered    only   positive   integral    exponents.     Such 
symbols  as  a3/4  and  or2  thus  have  no  meaning  for  us  as  yet 
since  there  can  be  no  such  thing  as  taking  a  as  a  factor  three 
fourths  times,  or  minus  two  times.     However,  we  shall  now  see 
that  by  extending  our  definitions  we  can  assign  perfectly 
definite  meanings  to  these  symbols  as  well  as  to  all  others 
wherein  fractional,  zero,  or  negative  exponents  occur. 

121.  Meaning  of  a  Fractional  Exponent.     If  a374  is  to 
obey  the  multiplication  law  (§  117)  then 


a         .  a         .  a         .  a        =  ^ 

That  is 

(a3/4)4  =  a3, 
so  that  we  must  have 


=     3/4+3/4+3/4+3/4  _ 


Thus  we  naturally  take  vV  to  be  the  meaning  of  a3/4. 
Similarly  (if  the  multiplication  law  is  to  hold  true),  the 
meaning  of  a273  is  v'a2,  while  that  of  a475  is  x/a4,  etc. 
So,  in  all  cases  am/n  means  the  nth  root  of  am,  that  is, 

Qinln  _  ^/Qm. 

Thus 

82/3  =  ^82  =  ^64  =  4. 
Similarly,  _         _ 

(2^4)8/4  =  #(x8g4)3  =  ^24012  =X*y3.         ($66  Rule  111    §   119.) 

EXERCISES 

Express  with  radical  sign  and  find  the  value  of  : 

1.  8173.  6.    27273.  11.    G/10)175- 

2.  41/2.  7.    32275.  12.    (i/10)275. 

3.  91/2.  8.    81374.  13. 

4.  27173.  9.    64176.  14. 

6.    (-8)173.  10.    (x6)173.  15.    S(a+6)3i273. 


XVIII,  §  123]  EXPONENTS  199 

Express  with  radical  signs  : 

16.  22/3.     18.  43/2.    20.   (a2)4/3.       22.  2  xl/\         24.  m2/3n3/4. 

17.  32/3.    19.  a473.    21.  (3z)3/4.     23.  (2  a&)3/4.     25.  (z+t/)5/6. 

Express  with  fractional  exponents  : 

26.  V^4.         29.  2v/z*~.  32.  Sv^n2".          35.  V(a+6)3. 

27.  V^5.         30.  ^(-a)2.      33. 

28.  v^S.        31.  Vabc.  34. 


122.  Meaning  of  a  Zero  Exponent.  If  a°  is  to  obey  the 
multiplication  law  (§117),  then  am-a°  =  am+°,  that  is 
am  •a°  =  am.  Dividing  both  members  of  the  last  equality 
by  am  gives  a°  =  a™  -f-  am  =  1  .  That  is, 


This  means  that  the  zero  power  of  any  number  a  (except  0) 
must  always  be  taken  equal  to  1. 

Thus  3°  =  1;  (-32)°  =  l;(£)o  =  l;xo  =  i;  (mn)°  =  l;  (o+6)°=l; 
etc. 

123.  Meaning  of  Negative  Exponents.  If  a~m  is  to  obey 
the  multiplication  law  (§117),  then  am  •  a~m  =  am-n  =  a°, 
that  is  am  •  a~m=l  (§  122).  Dividing  both  members  of  the 
last  equation  by  am  gives 


This  means  that  a  negative  power  of  any  number  a  must  al- 
ways be  taken  equal  to  1  divided  by  the  corresponding  positive 
power  of  a. 


2'     8'  (-4)'     1 

Similarly,  („+!>)-"*  =  — L—  —±=. 


200  SECOND   COURSE    IN   ALGEBRA     [XVIII,  §  123 

EXERCISES 

Express  with  positive  exponents  and  find  the  values  of 
each  of  the  following  expressions. 

1.  S-2.       5.  2-1  •  3-2.  9.  82  •  4~*.          13.  81~1/4. 

2.  4~2.       6.  4°  -  S-3.  10.  8~1/3.  14.  64~1/6. 

3.  2~4.       7.  7-4-4.  11.  (-8)-1/3.       15.  (-125)~1/3. 

4.  8°.        8.  2~3  -8  •4-».       12.  27~1/3.  16.  (-32)-1/5. 

Write  with  positive  exponents  each  of  the  following  ex- 
pressions. 

17.  x3y~*.  20.  (2  a)"3?)3.  23.  §~lmin~*. 

18.  x~2y2<r3.  21.  2~3a-363.  24.  (a26c)"2. 

19.  2a~363.  22.  (-m)-\-ri)-\  25.   \a?(m-n) \~\ 

124.  Negative  Exponents  in  Fractions.  This  is  best 
understood  from  an  example. 

EXAMPLE.     Write a   o    with  positive  exponents  only. 
cr2 

1.  53 

fl      O        d  O  O          Cr       u  C?  A  /c   i  oo\ 

SOLUTION.    !^f-™ — i —  =  —^          4  '  T=  ~T*     ^ns.          (§  126) 
<? 

It  is  to  be  observed  that  the  answer  here  results  directly 
by  transferring  the  factor  cr4  to  the  denominator  by  changing 
the  sign  of  its  exponent,  and  transferring  the  factor  c~2  to 
the  numerator  by  likewise  changing  the  sign  of  its  exponent. 

Thus  we  have  the  following  important  principle. 

A  factor  may  be  transferred  from  either  term  (numerator 
or  denominator)  of  a  fraction  to  the  other  provided  the  sign  of 
its  exponent  be  changed. 

Thus  we  may  write  x  ^~_^—  =  -^ — -. 

Similarly,  4  q'^c"a=  4  a3b-3c-^d*e-6.  Here  we  have  written  the 
fraction  in  a  form  having  no  denominator,  that  is  as  a  product. 


XVIII,  §  125J  EXPONENTS  201 


EXERCISES 


Write  each  of  the  following   expressions   with   positive 
exponents  only. 

a-*b  3  a-3 


c  '   (26) 


y-l     ' 


-2 


2(c+d)e-4 

Change  each  of  the  following  expressions  to  the  form  of  a 
product. 

8 


'  " 


11    JL2.       13    3r3g'2         15 
'  '  ' 


125.  The  Fundamental  Laws  for  Any  Rational  Exponent. 

The  five  fundamental  laws  stated  in  §  117  were  there  proved 
true  only  for  positive  integral  exponents,  but  it  can  be  shown 
that  they  hold  equally  well  for  fractional,  zero,  or  negative 
exponents.  As  the  proof  of  this  fact  is  long,  it  will  be 
omitted  from  this  text.  The  following  illustrations  of  the 
meaning  of  the  laws  in  such  cases  should,  however,  be  care- 
fully examined. 


1  .   a6  •  a-3  •  a4/3  •  a273  =  a*-3+473+2/3  =  a6-^2  =  a5.     (Law  I) 


2.         =  a5/6-(-3)  =  a5/6+3  =  a¥  =  a3|>  (Law  II) 

cr3 


3.  (a-372)475  =  a'-372)  '  4/5  =  a-675.  (Law  III) 

4.  (a-^62)-1/4  =  a(-3)(-174)  •  62<-1/4>  =  a374&-172.  (Law  IV) 


202  SECOND   COURSE    IN   ALGEBRA     [XVIII,  §  125 

HISTORICAL  NOTE.  The  idea  of  using  exponents  to  mark  the 
power  to  which  a  quantity  is  raised  is  due  to  the  French  mathe- 
matician and  philosopher  Descartes  (1596-1650)  ;  see  the  picture  fac- 
ing p.  41),  but  he  used  only  positive  integral  exponents,  as  in  a1, 
a2,  a3,  a4,  •••.  The  English  mathematician  Wallis  (1616-1703)  en- 
couraged the  use  of  fractional  and  negative  exponents  and  caused 
them  to  be  brought  into  general  use. 

EXERCISES 

In  the  following  exercises,  assume  that  the  laws  of  §  117 
hold  true  for  all  rational  exponents. 

LAW  I 

Multiply 

1.  a2  by  a-1.  8.  a1/261/3  by  a1/262/3, 

2.  a3  by  a~2.  9.  ra1/2rc1/3  by  m3/2rc2/3. 

3.  a3  by  or3.  10.  p~1/4  q  by  4  p5/4. 

4.  a  by  cr4.  11.  n2  by  6n~3. 

5.  cr2  by  a-3.  12.  xm~n  by  xm+n. 

6.  a173  by  a2/3.  13.  z(m+n)/2  by  x(m~n)/2. 

7.  ar1/2byz3/2.  14.  a1/2+61/2  by  a1/261/2. 

16.   a2/3+a1/361/3+&2/3  by  a1/3-61/3. 

[HINT.  Follow  the  rule  for  multiplying  one  polynomial  by 
another,  as  given  in  §  8.] 

16.  m1/3+m1/6n1/6+n1/3  by  m1/3-m1/6n1/6+n1/3. 
Carry  out  the  following  indicated  operations. 

17.  (a1/2+61/2)(a1/2-61/2).          20. 

18.  (x2/3+2/2/3)(x2/3-i/2/3).          21. 

19.  (a1/2+61/2)2.  22. 
23.    (z2 

24.  (x2 

25.  (2z2/3-3z1/3+4)(2+3ar1/3). 

26.  x2- 


XVIII,  §  125]  EXPONENTS  203 

LAW  II 

Divide 

27.  a4  by  a5.      29.   z2  by  or2.  31.    (ran)2/3  by  (mnY/s 

28.  a3bya°.      30.   z372  by  z~1/2.      32.   xl/  Y/2  by  y~l/2. 

33.  x*-\-x2y2-\-y*  by  z2?/2. 

SOLUTION.     x  ~*~x  ^  +3r=_g — |_?J£.  _| — ^_=^_. 

X2^2  X2^2        X2^2        X2^2        ^/2 

34.  a3+a2+abya4. 

35.  a~4+a~26+62  by  a~2b. 

36.  x4-|-2  az3-|-5  z"1?/  —  Q?]r^~}~y*  by  x—12/~2. 

37.  a  — 6  by  a1/2+61/2. 
SOLUTION,     a  —61 


a+a1/2b1/2          a1/2-61/2. 
— a1/261/2  — 6 


38.  a-6bya1/2-61/2.  41.   z-1  by  z2/3+z1/3+l. 

39.  a-f6  by  a1/3+61/3.  42.   x  -  y  by  z174  - 1/1/4. 

40.  z2+2/2  by  z2/3+2/2/3.  43.   m2-n3  by  w1/3+n1/2. 

LAW  III 

Simplify 

44.  (a1/2)3. 

SOLUTION.     As  in  Illustration  3  of  §  125,  the  answer  is  a1/2X3, 
or  a3/2. 

45.  (a1/2)2.  46.  (or173)6.  47.  (or5)2.  48."(8-1/3)2.  49.  (16~1/2)3. 
50.    Var2.        [HINT.     V^  =  (a~2)1/2  by  §  121.] 

61.    VaF"2.  52.    v^z372. 


204  SECOND   COURSE    IN   ALGEBRA     [XVIII,  §  125 

LAWS  IV  AND   V 
Simplify 

63. 


SOLUTION.    As  in  Illustration  4  of  §  125,  we  have  (a~462/3)-1/2  = 

4)(-l/2)  .    52/3(-l/2)  =  a25-l/3. 


64.  (al73&-172)6. 

65.  (z4732T3)-173. 

66.  vV1726-3. 
57.  Vz473?/-3. 


58. 
59. 
60. 
61. 


[HINT  to  Ex.  62.     See  Illustration  5  in  §  125.] 
63.  .  64. 


*  MISCELLANEOUS   EXERCISES 

Expand,  by  use  of  Formulas  VI  and  VII  of  §  10. 

1.  (a:172-i/172)2.  3.    (a173+6173)2. 

2.  (a^+Zr1)2.  4.    (1+2  x172)2. 
Simplify,  expressing  results  with  positive  exponents. 


6. 


-VWxVa-* 


z  y 


^T^/W"  10. 

Solve  forz  and  check  each  result  in  the  following  equations. 

11.  z3/4  =  8.     [HiNT.    Write  z3'4  in  the  form  Or1/4)3.] 

12.  z275  =  9.  14.   £z372  =  72.          16.   25ar273  =  l. 

13.  z473  =  16.  16.   la;273  =  25.          17.   or372 -27  =  0. 


CHAPTER  XIX 
RADICALS 

126.  Important  Formulas.  In  §  3  we  defined  VcTas  mean- 
ing that  number  or  expression  which,  when  raised  to  the  nth 
power,  would  give  a  ;  that  is 

(1)  (W  =  a. 

Unless  a  is  an  exact  nth  power  of  some  number  or  expres- 
sion, we  agreed  (§  41)  to  call  Va  a  radical  of  the  nth  order. 

Thus  V5,  V23,  Vj,  V.05,  Vx+y,  Vm2+n2  are  radicals  of  the 
second  order;  #5~,  v^  ^f'  v/x+y  are  radicals  of  the  third  order, 
etc. 

Moreover,  we  saw  (§  44)  that  there  exist  two  general 
formulas  as  follows  : 

(2) 

_  Va 
(3) 


And,  in  connection  with  the  study  of  fractional  exponents, 
we  havejseen  (§  121)  that  the  meaning  of  a1/n  must  be  taken 
to  be  Va,  that  is  we  have  the  formula 

(4)  a1/n=Va. 

The  four  formulas  (1),  (2),  (3),  (4)  contain  all  that  is 
essential  in  the  study  of  radicals.  In  fact,  we  have  already 
seen  in  Chapter  IX  how  (1),  (2),  and  (3)  are  thus  used.  In 
the  present  chapter  we  shall  review  and  extend  those  studies, 
making  use  now  of  (4)  also. 

205 


206 


SECOND   COURSE    IN   ALGEBRA     [XIX,  §  127 


127.   Simplification  of  Radicals. 

EXAMPLE  1.     Simplify  V75. 

SOLUTION.     V75  =  V25X3  =  V25  X  Vjj  =  5  Vs.     Ans. 

(Formula  (2),  §  126) 

Ay-J     A  fit    *T 

EXAMPLE  2.     Simplify 
SOLUTION. 


Ans. 


EXERCISES 

Before  undertaking  the  following  exercises,  review  §  44,  including 
the  note  on  page  72.  These  resemble  the  exercises  on  pages  73,  74, 
but  in  some  instances  are  more  difficult  because  they  refer  to  radicals 
of  as  high  orders  as  the  fifth,  sixth,  and  seventh. 

Simplify  each  of  the  following  expressions. 


3/3.    ^3 

\«*        9 


29. 


30.    A  — 


31. 


27 


32. 


33. 


34. 


8/  4 
\125 


[HINT.     See  Note  in  §  45.] 
35. 


81 


37. 


XIX,  §  128]  RADICALS  207 

EXERCISES 

Write  each  of  the  following  in  a  form  having  no  coefficient 
outside  the  radical  sign.  First  review  the  similar  exercises 
on  page  74. 

1.  3^2. 

SOLUTION.     3^2  =  ^27X^2  =  v/27><2  =  ^54.     Ans. 

2.  2\/3.  7.   4oV2a.  12. 

3.  2\/2.  8.    GsVjfi?.  2r 

13     ^ 

4.  3A/3.  9. 


5.  2"V/f.  10.   mv^n.  2x 

6.  3^6.  11.   2a^.  T    2 

128.  Reduction  of  Radicals  of  Different  Orders  to  Equiva- 
lent Radicals  of  the  Same  Order. 

EXAMPLE.  Reduce  V%  v^,  and  ^5  to  equivalent  radi- 
cals of  the  same  order. 

SOLUTION.  By  use  of  Formula  (4),  §  126,  and  the  laws  of  expo- 
nents (§  117)  we  may  write 

V2"=  21/2  =  26/12  =  v^»  =  V64, 

#3=  31/3  =  34/12  =  v^3*  =  v/81, 
^5  =51/4  =53/12  =  ^5?=  1^ 


NOTE.  As  now  expressed,  the  given  radicals  may  be  compared 
as  to  their  magnitudes.  Thus  we  see  V5  is  the  greatest  of  the 
three  since  (the  orders  of  the  radicals  being  now  the  same)  it  has  the 
largest  radicand,  namely,  125. 

An  examination  of  the  process  just  followed  leads  to  the 
following  rule. 

To  reduce  radicals  to  equivalent  radicals  of  the  same  order  : 

1.  Express  the  radicals  with  fractional  exponents. 

2.  Reduce  the  exponents  to  a  common  denominator. 

3.  Rewrite  the  results  thus  obtained  in  radical  form. 


208 


SECOND    COURSE    IN   ALGEBRA     [XIX,  §  128 


EXERCISES 

Reduce  each  of  the  following  groups  to  equivalent  radicals 
of  the  same  order. 

1.  V3andV?.  3.    Vjfand  V<T 

2.  V2  and  Vs.  4.    V5,  V6,  and 

5.  Vo~and  VS.         Va3  and  VP".  Ans. 

6.  V2~i  and  \^3x. 

7.  Vo-?/,  V?/2,  and  Vxz. 
8. 

9. 


Arrange  the  following  in  order  of  magnitude.     (See  §  128.) 

11.  V3,  -^4.  14.    VI,  Vl3. 

12.  V2,  V3.  16.    V2,  V4,  V6. 

13.  VlO,  V5.  16.    Vl3,  V7,  vT74. 

129.  Multiplication  of  Radicals.  We  have  already  seen 
in  §  46  how  to  multiply  two  or  more  radicals  of  the  second 
order.  Radicals  of  higher  order  than  the  second  may  be 
multiplied  by  means  of  the  general  formula  (2)  of  §  126. 

EXAMPLE  1.  VI-  VIO  =  V40.  (Formula  (2),  §126.)  Sim- 
plifying (§  127),  V40  =  2VB.  Ans. 

EXAMPLE  2.    VTx  - 


But 


=  V(4z)3  •  (2z2)2.     (Formula  (2),  §  126) 


V(4  a 


2)2=  V43  •  22  •  a;3  •  x*=  V26  •  22  •  x1 
=  V(2  x)6  -  (4  x)  =  2  x  V4z.     Ans. 

Note  that  in  Ex.  2  the  given  radicals  are-  of  different  orders, 
in  which  case  the  first  step  is  to  reduce  them  to  equivalent  radicals 
of  the  same  order  (§  128).  Then  apply.  (2)  of  §  126. 


XIX,  §  130]  RADICALS  209 

In  general,  we  have  the  following  rule. 
To  multiply  radicals  of  any  orders  : 

1.  Reduce  the  radicals,  if  necessary,  to  equivalent  radicals 
of  the  same  order  (§  128). 

2.  Multiply  the  resulting  radicals  by  use  of  Formula  (2), 
§126. 

3.  Simplify  the  result  as  in  §§  44  and  127. 

EXERCISES 

[Compare  with  the  exercises  on  page  76.] 
Find  each  of  the  following  indicated  products. 


We  have  already  seen  in  §  47 
how  one  radical  of  the  second  order  may  be  divided  by  another 
of  that  order.  If  the  radicals  are  of  higher  order  than  the 
second,  we  may  divide  them  by  means  of  the  general  formula 
(3)  of  §  126. 
EXAMPLE  1.  Divide  vT6  by 


SOLUTION.  =          =  (Formula  (3),  §126) 


The  simplest  and  most  desirable  form  in  which  to  leave  this 
result  is  that  in  which  no  fraction  appears  except  outside  the 
sign.  Thus 


210  SECOND    COURSE    IN   ALGEBRA     [XIX,  §  130 

EXAMPLE  2.     Divide  V3  by  \/9. 

\i«          (Formula  (3),  §  126) 


But  ^^E-iSLl* 

EXAMPLE  3.     Divide  \^3xyby  \/3  X2y*. 

SOLUTION.     -E  =  ^^  =  ^^=  $• 
V3r.2?y3       -v/s  T2,;3        *3  o;2-?;3       "ti 

But 


In  general,  we  have  the  following  rule. 
To  divide  one  radical  by  another  : 

1.  Reduce  the  radicals,  if  necessary,  to  equivalent  radicals 
of  the  same  order  (§  128). 

2.  Divide  the  resulting  radicals  by  use  of  Formula  (3)  (§  126). 

3.  Simplify  the  result  in  such  a  way  that  no  fraction  appears 
except  outside  the  radical  sign. 

EXERCISES 

[Compare  with  the  exercises  on  page  77.] 
Find  each  of  the  following  indicated  quotients. 

1.    V7  +  V2.          4.   2  +  \/2.          7.    \/l2x2+V2~x. 

5.    3-^3.          8.    v^81  xzy+\/9~xy. 

-r-  ^32  m. 


9. 


13. 
14. 
15. 


XIX,  §  132]  RADICALS  211 

131.  Involution  and  Evolution  of  Radicals.  By  use  of 
Formula  (4)  of  §  126  together  with  the  general  laws  of  expo- 
nents (§  117),  we  may  raise  a  radical  to  a  power  or  extract 
a  root  of  it. 

EXAMPLE  1.     Find  the  square  of  \/8. 

SOLUTION.     ( v'S)2  =  (81/4)2  =82/4  =81/2  =  V8  =2  V2.     Ans. 

EXAMPLE  2.     Find  the  fourth  root  of  V2  x. 

SOLUTION.      ^  V2x  =  [(2  x)1/2]1/4  =  (2  zW*  =  \/2aT.     Arts. 

EXERCISES 

Perform  each  of  the  following  indicated  involutions. 

1.  (v/3)2.  6.    (2V3)3. 

2.  (v/S)2.  7.    (3V2)3. 

3.  (3Vi)2.  8. 

4.  (2^3l02.          9. 

5.  x2vT2.       10. 


Perform  each  of  the  following  indicated  evolutions. 

16.  ^V2. 

17.  V^f. 

is.  v^p; 

19.    V</|9. 

132.  Rationalizing  the  Denominator  of  a  Fraction.     If  the 

denominator  of  a  fraction  consists  of  a  single  quadratic  radi- 
cal, or  is  a  binomial  containing  quadratic  radicals,  the  frac- 
tion may  be  changed  into  one  which  has  radicals  only  in  its 
numerator.  The  process  of  doing  this  is  called  rationalizing 
the  denominator. 


212  SECOND   COURSE    IN   ALGEBRA     [XIX,  §  132 

EXAMPLE  1.     Rationalize  the  denominator  in  the  fraction 

Vf 
VB 

SOLUTION.  Here  the  denominator  contains  the  single  surd  V5. 
To  rationalize  this  denominator  it  is  merely  necessary  to  multiply 
both  numerator  and  denominator  by  V5^  giving  Vl5/5.  Ans. 

EXAMPLE  2.     Rationalize  the  denominator  in  the  fraction 


V3+V2 

SOLUTION.  Multiply  both  numerator  and  denominator  by 
V3  —  V2,  giving 

V3-V2  =      V3-V2     ^  V3-  V2  =  V3-  V2 

(V3-V2)(V3  +  V2)     (V3)2-(V2)2        3-2  1 

=  V3-V2.     Ans. 

EXAMPLE  3.     Rationalize  the  denominator  in  the  fraction 

3V5+2V2 
V5-V2 

SOLUTION.  Multiplying  both  numerator  and  denominator  by 
V5  +  V2,-  we  have 

=  3-  5+3V10+2V10+2  •  2 
5-2 

=  19+|V10     AnSf 

We  may  then  state  the  following  rule. 

To  rationalize  the  denominator  of  a  fraction  : 

If  the  denominator  contains  a  single  radical,  multiply  both 
numerator  and  denominator  by  that  radical. 

If  the  denominator  has  either  of  the  binomial  forms  Va+  Vb 
or  Va—  VS,  multiply  both  numerator  and  denominator  by 
Va  —  \/b,  or  Va-j-VS  according  as  we  have  the  first  or  second 
of  these  two  cases. 


XIX,  §  133]  RADICALS  213 


EXERCISES' 


Rationalize  the  denominators  in  each  of  the  following 
fractions. 


5    V3+V2  9    3V3-2V2 

VI  V3-A/2  '  2V3+3\/2 

6-    -7F— '  10. 


2\/5 
7.   -  —  -.  -    2\/q-3\/b 


s.  j 

133.   Finding  the  Value  of  Fractions  Containing  Radicals. 
Suppose  we  wish  to  find  the  value  of 


V3+V2 

correct  to  five  places  of  decimals.  It  is  well  to  begin  by 
rationalizing  the  denominator,  thus  making  the  fraction  take 
the  form  (see  Ex.  2  worked  in  §  132)  \/3  -  \/2.  All  we  now 
need  to  do  is  to  look  up  in  the  table  the  values  of  V3  and  V2 
so  as  to  work  out  the  value  of  A/3  —  \/2.  That  is,  we  have 


=V3-\2  =  1.73205+-1.41421+  =  0.31784+.    Ans. 


V3+V2 

If,  in  this  example,  we  had  not  first  rationalized  the  denomi- 
nator, we  should  have.  had  to  find  the  value  of 

1 


or 


1.73205++ 1.41421+'  3.14626+' 

which  would  compel  us  to  divide  1  by  3.14626.  Note  how 
much  more  difficult  this  is  than  the  above,  where  virtually 
all  we  need  to  do  is  to  subtract  1.41421  from  1.73205. 


214  SECOND   COURSE    IN  ALGEBRA     [XIX,  §  133 

This  illustrates  the  general  fact  that  to  find  the  value  of  a 
fraction,  its  denominator  (if  it  contains  radicals)  should  first 
be  rationalized  whenever  possible. 

EXERCISES 

Find  (by  first  rationalizing  the  denominator  and  then  using 
the  table)  the  approximate  values  of  the  following  fractions. 

2  ,    1_  ,         1 

1.    — — .  3.    — — —  •  5. 


V3  V3-\/2  V250 

2\/5-4 


3V5  2-V3  '   3V3-2 

*134.  Binomial  Surd.  A  binomial,  one  or  both  of  whose  terms 
are  surds  (§  42),  is  called  a  binomial  surd. 

Thus  2  + A/5,  A/2  +  A/5,  A/3-1  are  binomial  surds. 

*135.  To  Find  the  Square  Root  of  a  Binomial  Surd.  The  famil- 
iar formula  for  (a  +6)2  (§  10,  Formula  VI)  may  be  put  into  the  form 

(1)  a2+62+2  ab  =  (a+&)2. 

Since  this  relation  holds  true  for  any  values  of  a  and  6,  let  us  sup- 
pose in  particular  that^both  are  positive,  in  which  case  we  may 
write  a  =  A/Z  and  b  =  Vy,  where  x  and  y  are  properly  chosen  positive 
values.  The  equation  just  written  then  takes  the  form 


Extracting  the  square  root  of  both  members  now  gives 

(2)  V 


This  formula,  having  been  thus  derived  from  (1),  must  therefore 
hold  true  for  any  positive  values  of  x  and  y. 

Similarly,  by  starting  with  the  familiar  formula  for  (a  —  6)2,  we 
arrive  at  the  formula 


(3) 

Formulas  (2)  and  (3)  are  frequently  used  to  obtain  the  square 
root  of  a  binomial  surd  (§  134)  as  illustrated  in  the  following 
example. 


XIX,  §  135]  RADICALS  215 

EXAMPLE.     Find  the  square  root  of  the  binomial  surd  11  +4  V7- 
SOLUTION.     We  are  to  find  V'n  +4  V?. 

This  may  be  written  Vn  +2  V28,  and  it  is  now  in  the  form  of  the 
first  member  of  Formula  (2),  provided  we  choose  x  and  y  so  that 
x  +y  =  11  while  xy  =  28. 

The  values  of  x  and  y  which  satisfy  these  last  two  equations  are 
seen  (by  inspection)  to  be  x  =  4,  y  =  7. 

Substituting  these  values  of  x  and  y  in  the  second  member  of 
(2)  gives  V4  +  V7=2  +  V7. 

Therefore  V^H  +4  V?  =2  +  V?.     Ans. 

CHECK. 
(2  +  V7)2=22+2-  2 


The  pupil  will  observe  that  all  that  is  essential  in  working  the 
above  example  is  to  write  the  given  surd  term  (4V7)  so  that  it  has 
the  coefficient  2  instead  of  4.  Similarly,  all  such  problems  may  be 
brought  under  Formulas  (2)  or  (3)  as  soon  as  the  coefficient  of  the 
surd  term  has  been  reduced  to  2. 

*  EXERCISES 

Find  the  square  root  of  each  of  the  following  expressions,  and 
check  your  answer. 

1.  6+2V8. 

[HINT.     Herez+?/=6,  xy  =8.] 

2.  6-2V8".       4.    ll-2V3a      6.    6  +  V32.         8.   8+4>/3. 

3.  7  +4  Vs.       5.    6-V2a          7.    7-V|a         9.   20-6VIT. 

10.  Establish  Formula  (3)  of  §  135  by  a  process  similar  to  that 
used  in  establishing  Formula  (2). 


CHAPTER  XX 
LOGARITHMS 

I.   GENERAL  CONSIDERATIONS  f 

136.  Definition  of  Logarithms.  If  we  ask  what  power 
of  10  must  be  used  to  give  a  result  of  100,  the  answer  is  2 
because  102=100.  Another  common  way  of  stating  this  is 
to  say  that  "  the  logarithm  of  100  is  2."  In  the  same  way, 
the  power  of  10  needed  to  give  1000  is  3  because  103  =  1000, 
and  this  is  briefly  stated  by  saying  that  "  the  logarithm  of 
1000  is  3."  Similarly,  the  power  of  10  that  gives  .1  is  —1 
because  10~1=11jy,  or  .1  (§  123),  and  this  is  equivalent  to 
saying  that  "  the  logarithm  of  .1  is  —  1."  Likewise,  the  loga- 
rithm of  .01  is  -2.  Why? 

From  these  illustrations  we  readily  see  what  is  meant  by 
the  logarithm  of  a  number.  It  may  be  denned  as  follows : 

The  logarithm  of  a  number  is  the  power  of  10  required  to 
give  that  number. 

NOTE.  A  more  general  definition  will  be  given  in  §  151,  but  this 
is  the  one  commonly  used  in  practice. 

The  fact  that  the  logarithm  of  100  is  2  is  written  log  100  =  2. 
Similarly,  we  have  log  1000  =  3,  log  .1  =  - 1,  log  .01  =  -2,  etc. 

t  Parts  I  and  II  give  definitions  and  essential  theorems  which 
should  be  well  understood  before  Part  III,  which  describes  the  im- 
portant applications,  is  taken  up. 

216 


XX,  §  137]  LOGARITHMS  217 

EXERCISES 

1.   What    is  the  meaning  of    log   10000  ?     What   is  its 
value  f 
»  2.   What  is  the  value  of  log  .001?     Why? 

3.  What  is  the  value  of  log  .00001  ?     Why? 

4.  What  is  the  value  of  log  10? 

5.  What  is  the  value  of  log  1  ?     (See  §  122.) 

6.  As  a  number  increases  from  100  to  1000  how  does  its 
logarithm  change? 

7.  As  a  number  decreases  from  .1  to  .01  how  does  its 
logarithm  change?     Answer  the  same  as  the  number  goes 
from  .01  to  .001  ;  from  1  to  10  ;  from  1  to  1000. 

8.  Explain  why  the  following  are  true  statements  : 
(a)  log  100000  =  5.  (6)  log  .0001  =-4. 

(c)  logv^0  =  i. 

[HINT.     Remember  VlO  =  101/2. 

(d)  log  >^IO  =  i. 

(e)  log  ^100  =  1. 

[HINT.     Remember  \/100  =  v/102  =  102'3.     (§  121.)] 


137.  Logarithm  of  Any  Number.  Suppose  we  ask  what 
the  value  is  of  log  236.  What  we  are  asking  for  (see  defini- 
tion in  §  136)  is  that  value  which,  when  used  as  an  exponent 
to  10,  will  give  236  ;  that  is  we  wish  the  value  of  x  which 
will  satisfy  the  equation  10X  =  236.  This  question  resembles 
those  in  §  136,  but  is  different  because  we  cannot  immediately 
arrive  at  the  desired  value  of  x  by  mere  inspection.  All  we 
can  say  here  at  the  beginning  is  that  x  must  lie  somewhere 
between  2  and  3,  because  102=100  and  103=1000,  and  236 
lies  between  these  two  numbers.  In  order  to  find  z  to  a  finer 
degree  of  accuracy,  it  is  now  natural  to  try  for  it  such  values 


218  SECOND    COURSE    IN   ALGEBRA      [XX,  §  137 

as  2.1,  2.2,  2.3,  2.4,  2.5,  2.6,  2.7,  2.8,  and  2.9,  all  of  which  lie 
between  2  and  3.  The  result  (which  for  brevity  we  shall 
here  state  without  proof)  is  that  when  x  =  2.3  the  value  of  10* 
is  slightly  less  than  our  given  number  236,  while  if  we  take 
x=  2.4  the  value  of  10*  is  slightly  greater  than  236.  Thus  x 
lies  somewhere  between  2.3  and  2.4.  In  other  words,  the 
value  of  log  236  correct  to  the  first  decimal  place  (compare 
§  37)  is  2.3. 

It  is  now  natural,  if  we  wish  to  obtain  x  to  still  greater 
accuracy,  to  try  for  it  such  values  as  2.31,  2.32,  2.33,  2.34, 
2.35,  2.36,  2.37,  2.38,  and  2.39,  all  of  which  lie  between  2.3 
and  2.4.  The  result  (which  again  is  here  stated  without 
proof)  is  that  when  x  =  2.37  the  value  of  10X  is  slightly  less 
than  our  number  236,  while  if  we  take  #  =  2.38  the  value  of 
10*  is  slightly  greater  than  236.  This  means  that  the  second 
figure  of  the  decimal  is  7,  after  which  we  may  say  that  the 
value  of  log  236  correct  to  two  places  of  decimals  is  2.37. 

Proceeding  farther  in  the  same  manner,  it  can  be  shown 
that  when  x  =  2.372  the  value  of  10*  is  slightly  less  than  236, 
while  for  x  =  2.373  the  value  of  10*  is  slightly  greater  than  236. 
Thus  the  value  of  log  236  correct,  to  three  places  of  decimals  is 
2.372.  Similarly,  it  can  be  shown  that  the  number  in  the 
fourth  decimal  place  is  9,  and  this  is  as  far  as  it  is  necessary 
to  carry  out  the  process,  since  the  result  is  then  sufficiently 
accurate  for  all  ordinary  purposes. 

In  summary,  then,  we  have  log  236  =  2.3729,  this  value 
being  correct  to  four  places  of  decimals. 

NOTE.  It  thus  appears  that  logarithms  do  not  in  general  come 
out  exact,  though  they  do  so  for  such  exceptional  numbers  as 
100,  1000,  10,000,  .1,  .01,  etc.  (Compare  §  37.)  They  can  be  ex- 
pressed only  approximately,  yet  as  accurately  as  one  pleases  by 
carrying  out  the  decimal  far  enough.  In  this  respect  they  resemble 
such  numbers  as  V2.  \/2,  V3,  etc. 


XX,  §  138]  LOGARITHMS  219 

Other  examples  of  logarithms  are  given  below.  Note 
especially  the  decimal  part  of  each,  which  is  correct  to  four 
places. 

log  283  =  2.4518  log  196  =  2.2923  log  17  =  1.2304 

log  6  =  0.7782  log  3.410  =  0.5328         log  5.75  =  0.7597 

138.  Characteristic.  Mantissa.  We  have  seen  that  the 
logarithm  of  a  number  consists  (in  general)  of  an  integral 
part  and  a  decimal  part. 

Thus  log  236  =2.3729.  Here  the  integral  part  is  2  and  the  deci- 
mal part  is  .3729.  Similarly,  in  log  6  =  0.7782  the  integral  part  is  0, 
while  the  decimal  part  is  .7782. 

These  two  parts  of  every  logarithm  are  given  special  names 
as  follows : 

The  integral  part  of  a  logarithm  is  called  the  characteristic 
of  the  logarithm. 

The  decimal  part  of  a  logarithm  is  called  the  mantissa  of 
the  logarithm. 

Thus  the  characteristic  of  log  236  is  2,  while  its  mantissa  is 
.3729.  (See  above.)  Similarly,  the  characteristic  of  log  6  is  0,  while 
its  mantissa  is  .7782. 

EXERCISES 

1.  What  is  the   characteristic   of  log   100?    What  the 
mantissa?    Answer  the  same  questions  for  log  1000,  log  10, 
and  log  1. 

2.  What  is  the  characteristic  of  log  156  ? 
[HINT.     Note  that  156  lies  between  102  and  103.] 

3.  What  is  the  characteristic  of  log  276?   of  log  1376?   of 
log  97?  of  log  18?  of  log  5?  of  log  11?  of  log  14798-? 

4.  For  what  kind  of  number  can  one  tell  by  inspection 
both  the  characteristic  and  the  mantissa  of  its  logarithm? 
(See  §  136.) 


220  SECOND   COURSE    IN   ALGEBRA      [XX,  §  1I19 

139.   Further  Study  of  Characteristic  and  Mantissa.     We 
have  seen  (§  138)  that  log  236  =  2.3729,  which  is  the  same  as 
saying  that 
(1)  102-3729  =  236. 

Let  us  now  multiply  both  members  of  (1)  by  10.     The 
left  side  becomes  102-3729+1  or   103-3729  (§  117,  Law  I)  while 
the  right  side  becomes  2360.     That  is,  we  have   103-3729  = 
2360,  which  is  the  same  as  saying  that 
log  2360  =  3.3729 

If,  instead  of  multiplying  both  sides  of  (1)  by  10,  we  divide 
both  by  10,  we  obtain  in  like  manner  lO2-3729"1  =  23.6  (§  117, 
Law  II).  That  is,  we  have  101-3729  =  23.6,  which  is  the  same 
as  saying  that 

log  23.6  =  1.3729 

Finally,  if  we  divide  both  sides  of  (1)  by  102,  or  100,  we 
obtain  102-3729~2  =  2.36.  That  is,  we  have  10°-3729  =  2.36  which 
is  the  same  as  saying  that 

log  2.36  =  0.3729 

What  we  now  wish  to  do  is  to  compare  the  results  which  we 
have  just  been  obtaining,  and  for  this  purpose  they  are  ar- 
ranged side  by  side  in  a  column  below. 

log  2360  =  3.3729 
log  236  =  2.3729 
log  23.6=1.3729 
log  2.36  =  0.3729 

Note  that  the  mantissas  here  appearing  on  the  right  are 
all  the  same,  namely  .3729,  while  the  numbers  appearing 
on  the  left  (that  is,  2360,  236,  23.6,  and  2.36)  are  alike  except 
for  the  position  of  the  decimal  point,  that  is  they  contain 
the  same  significant  figures.  This  illustrates  the  following 
important  rule. 


(2) 


XX,  §  141]  LOGARITHMS  221 

RULE  I.  //  two  or  more  numbers  have  the  same  significant 
figures  (that  is,  differ  only  in  the  location  of  the  decimal  point) , 
their  logarithms  will  have  the  same  mantissas,  that  is  their 
logarithms  can  differ  only  in  their  characteristics. 

Thus  log  243,  log  2430,  log  24.3,  log  2.43,  log  .243,  and  log  .0243 
all  have  the  same  mantissas.  It  is  only  their  characteristics  that 
can  be  different. 

EXERCISES 

Apply  Rule  I,  §  139,  to  tell  which  of  the  following  loga- 
rithms have  the  same  mantissas. 

log  .167        log  8100        log  16.7        log  81  log  .0072 

log  .081         log  7.2  log  720         log  1670        log  16700 

II.  To  DETERMINE  THE  LOGARITHM  OF  ANY  NUMBER 

140.  Purpose  of  This  Part.     When  we  wish  to  determine 
the  value  of  a  logarithm,  as,  for  example,  to  find  log  236,  we 
can  work  out  the  characteristic  and  mantissa  as  explained 
in  §  137,  but  this  requires  considerable  time.     What  we  do 
in  practice  is  to  use  certain  simple  rules  for  determining  the 
characteristic,  and  we  determine  the  mantissa  directly  from 
certain  tables  which  have  been  carefully  prepared  for  the 
purpose.     We   shall   now   state    these    rules    (§§  141-143) 
and  explain  the  tables  and  how  to  use  them  (§§  144-146). 

141.  Characteristics  for  Numbers  Greater  than  1.    If  we 
look  again  at  the  results  in  (2)  of  §  139,  we  see  that  the 
characteristic  of  log  2360  is  3.     Thus  the  characteristic  is  1 
less  than  the  number  of  figures  to  the  left  of  the  decimal 
point. 

NOTE.  2360  is  the  same  as  2360.,  so  that  there  are  four  figures 
here  to  the  left  of  the  decimal  point. 


SECOND   COURSE    IN   ALGEBRA      [XX,  §  141 

Again,  we  see  from  (2)  of  §  139  that  the  characteristic  of 
log  236  is  2  and  this,  as  in  the  case  already  examined,  is  1  less 
than  the  number  of  figures  to  the  left  of  the  decimal  point. 

NOTE.  236  is  the  same  as  236.,  so  there  are  three  figures  here  to 
the  left  of  the  decimal  point. 

Similarly,  since  the  characteristic  of  log  23.6  is  1  (see  (2) 
of  §  139)  this  again  obeys  the  same  law  as  just  observed  in 
the  other  two  cases,  that  is,  the  characteristic  is  1  less  than  the 
number  of  figures  to  the  left  of  the  decimal  point. 

Finally,  since  the  characteristic  of  log  2.36  is  0,  the  same 
law  is  again  present  here.  Explain. 

The  law  which  we  have  just  observed  can  be  shown  in  like 
manner  to  hold  good  for  the  characteristic  of  the  loga- 
rithm of  any  number  greater  than  1 ;  hence  we  may  state 
the  following  general  rule. 

RULE  II.  The  characteristic  of  the  logarithm  of  a  number 
greater  than  1  is  one  less  than  the  number  of  figures  to  the  left 
of  the  decimal  point. 

Thus  the  characteristic  of  log  385.9  is  2 ;  that  of  log  8.679  is  0. 
EXERCISES 

State,  by  Rule  II,  §  141,  the  characteristic  of  the  logarithm 
of  each  of  the  following  numbers. 

1.  476.5  5.  89.65  9.  500.005 

2.  325.  6.  105,000.  10.  3076.8 

3.  8976.  7.  17.694  11.  41. 

4.  1.6  8.  2.0815  12.  3.25679 

State  how  many  figures  precede  the  decimal  point  of  a 
number  if  the  characteristic  of  its  logarithm  is 

13.   3.        14.    2.        15.   0.        16.    1.        17.   4.        18.   5. 


XX,  §  142]  LOGARITHMS  223 

142.   Characteristics  for  Positive  Numbers  Less  Than  1. 
We  have  seen  (see  (2)  in  §  139)  that  log  2.36  =  0.3729,  which 
is  the  same  as  saying  that 
(1)  10°-3729  =  2.36 

Let  us  now  divide  both  members  of  this  relation  by  10. 
We  thus  obtain  (§  117,  Law  II) 

10o.3729-i  =  .236      (or  10-1+0-3729  =  .236) , 
which  gives  us  (by  §  136) 

log  .236  =-1+0.3729 

Observe  that  —  1  +0^3729  is  really  a  negative  quantity,  being 
equal  to  -(1-0.3729)  which  reduces  to  -0.6271.  However,  it  is 
more  convenient  for  our  present  purposes  to  keep  the  longer  form 
—  1+0.3729.  Note  that  this  cannot  be  written  as  —1.3729  be- 
cause this  last  is  equal  to  -1-0.3729  instead  of  -1+0.3729. 

If,  instead  of  dividing  both  members  of  (1)  by  10,  we 
divide  both  by  102,  or  100,  we  obtain 

JQO.3729-2  =  <0236        (or  lQ-2+0.3729  =  .Q236), 

which  means  that 

log  .0236  =-2+ 0.3729 

Similarly,  by  dividing  (1)  by  103,  or  1000,  we  find  that 

log  .00236=  -3+0.3729 

Finally,  if  we  divide  (1)  by  104,  or  10000,  we  find  that 
log  .000236=  -4+0.3729 

Let  us  now  compare  the  four  results  just  obtained.  Be- 
ginning with  the  last  result,  we  see  that  in  the  number 
.000236  there  are  three  zeros  immediately  to  the  right  of  the 
decimal  point,  that  is,  between  the  decimal  point  and  the 
first  significant  figure.  Corresponding  to  this,  the  charac- 
teristic on  the  right  is  minus  four.  Hence  the  characteristic 
is  negative  and  1  more  numerically  than  the  number  of  zeros 
between  the  decimal  point  and  the  first  significant  figure. 


224  SECOND   COURSE    IN   ALGEBRA      [XX,  §  142 

Similarly,  in  the  number  .00236  there  are  two  zeros  between 
the  decimal  point  and  the  first  significant  figure,  and  corre- 
sponding to  this  there  is  a  characteristic  on  the  right  of 
minus  three.  Hence,  as  before,  the  characteristic  here  is 
negative  and  numerically  1  more  than  the  number  of  zeros 
between  the  decimal  point  and  the  first  significant  figure. 
This  statement,  which  is  true  in  all  cases  mentioned  above,  can 
be  proved  for  the  characteristic  of  the  logarithm  of  any 
positive  number  less  than  1.  Hence  we  have  the  following 
rule. 

RULE  III.  The  characteristic  of  the  logarithm  of  a  (positive) 
number  less  than  1,  is  negative,  and  is  numerically  1  greater 
than  the  number  of  zeros  between  the  decimal  point  and  the  first 
significant  figure. 

Thus  the  characteristic  of  log  .0076  is  -3  ;  that  of  log  .28  is  -1. 

NOTE.  The  logarithm  of  a  negative  number  is  an  imaginary 
quantity  (as  shown  in  higher  mathematics),  and  hence  we  shall  con- 
sider here  the  logarithms  of  positive  numbers  only. 

143.  Usual  Method  of  Writing  a  Negative  Characteristic. 
In  §  142  we  saw  that  log  .236=  -1+0.3729.  If  we  add  10 
to  this  quantity  and  at  the  same  time  subtract  10  from  it, 
we  do  not  change  its  value,  but  we  give  it  the  new  form 
9+0.3729-10,  which  is  the  same  as  9.3729-10.  That  is, 
we  may  write. 

log  .236  =  9.3729 -10. 

This  is  the  form  used  in  practice. 

Likewise,  instead  of  writing  log  .0236= -2+0.3729  (see 
§  142)  we  write  in  practice 

log  .0236  =  8.3729-10, 
and  similarly  we  write 

log  .00236  =  7.3729 -10. 


XX,  §  144]  LOGARITHMS  225 

Thus  the  usual  method  of  expressing  the  characteristic 
of  —  1  is  to  write  9—10  for  it ;  if  it  is  —  2,  we  write  8  — 10  for 
it ;  if  it  is  —3,  we  write  7  —  10  for  it,  etc. 

For  example,  log  .0076  has  the  characteristic  7  — 10. 

EXERCISES 

State,  by  Rule  III,  §  142,  the  value  of  the  characteristic 
of  the  logarithm  of  each  of  the  following ;  state  how  it  would 
be  written  if  expressed  in  the  usual  form  described  in  §  143. 

1.  .06      -2,  or  8-10.     Ans. 

2.  .0087  5.    .0835  8.    .00978 

3.  .75  6.    .835  9.    .12345 

4.  .00067  7.    .33764 

How  many  zeros  lie  between  the  decimal  point  and  the  first 
significant  figure  of  a  number  when  the  characteristic  of  its 
logarithm  is 

10.    -3.     11.    9-10.     12.    -5.     13.   8-10.     14.    7-10. 

144.  Determination  of  Mantissas.  Use  of  Tables.  Sup- 
pose we  wish  to  determine  completely  the  value  of  log  187. 
By  Rule  II,  §  141,  we  know  that  the  characteristic  is  2. 
To  find  the  mantissa,  we  turn  to  the  tables  (p.  226)  and 
look  in  the  column  headed  N  for  the  first  two  figures  of  the 
given  number,  that  is,  for  18.  The  desired  mantissa  is  then 
to  be  found  on  the  horizontal  line  with  these  two  figures  and 
in  the  column  headed  by  the  third  figure  of  the  given  num- 
ber, that  is,  in  the  column  headed  by  7.  Thus  in  the  present 
case  the  mantissa  is  found  to  be  .2718. 

NOTE.  For  brevity,  the  decimal  point  preceding  each  mantissa 
is  omitted  from  the  tables.  It  must  be  supplied  as  soon  as  the 
mantissa  is  used. 

The  complete  value  (correct  to  four  decimal  places)  of  log 
187  is  therefore  2.2718. 


226 


SECOND   COURSE   IN  ALGEBRA 


N 

O 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1O 

11 
12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 
1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 
1614 

0253 
0645 
1004 
1335 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1072 
1399 
1703 

0374 
0755 
1106 
1430 
1732 

15 

16 
17 
18 
19 

1761 
2041 
2304 
2553 

2788 

1790 
2068 
2330 
2577 
2810 

1818 
2095 
2355 
2601 
2833 

1847 
2122 
2380 
2625 
2856 

1875 
2148 
2405 
2648 
2878 

1903 
2175 
2430 
2672 
2900 

1931 
2201 
2455 
2695 
2923 

1959 

2227 
2480 
2718 
2945 

1987 
2253 
2504 
2742 
2967 

2014 
2279 
2529 
2765 
2989 

2O 

21 
22 
23 
24 

3010 
3222 
3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 
3838 

3075 
3284 
3483 
3674 
3856 

3096 
3304 
3502 
.3692 
3874 

3118 
3324 
3522 
3711 

3892 

3139 
3345 
3541 
3729 
3909 

3160 
3365 
3560 
3747 
3927 

3181 
3385 
3579 
3766 
3945 

3201 
3404 
3598 
3784 
3962 

25 

26 
27 
28 
29 

3979 
4150 
4314 
4472 
4624 

3997 
4166 
4330 
4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216 
4378 
4533 
4683 

4065 
4232 
4393 
4548 
4698 

4082 
4249 
4409 
4564 
4713 

4099 
4265 
4425 
4579 
4728 

4116 
4281 
4440 
4594 
4742 

4133 
4298 
4456 
4609 
4757 

30 

31 
32 
33 
34 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
5366 

4843 
4983 
5119 
5250 
5378 

4857 
4997 
5132 
5263 
5391 

4871 
5011 
5145 
5276 
5403 

4886 
5024 
5159 
5289 
5416 

4900 
5038 
5172 
5302 
5428 

35 

36 
37 
38 
39 

5441 
5563 
5682 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 
5587 
5705 
5821 
5933 

5478 
5599 
5717 
5832 
5944 

5490 
5611 
5729 
5843 
5955 

5502 
5623 
5740 

5855 
5966 

5514 
5635 
5752 
5866 
5977 

5527 
5647 
5763 
5877 
5988 

5539 
5658 
5775 
5888 
5999 

5551 
5670 
5786 
5899 
6010 

40 
41 
42 
43 
44 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 
6444 

6042 
6149 
6253 
6355 
6454 

6053 
6160 
6263 
6365 
6464 

6064 
6170 
6274 
6375 
6474 

6075 
6180 
6284 
6385 
6484 

6085 
6191 
6294 
6395 
6493 

6096 
6201 
6304 
6405 
6503 

6107 
6212 
6314 
6415 
6513 

6117 
6222 
6325 
6425 
6522 

45 

46 

47 
48 
49 

6532 
6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6561 
6656 
6749 
6839 
6928 

6571 
6665 
6758 
6848 
6937 

6580 
6675 
6767 
6857 
6946 

6590 
6684 
6776 
6866 
6955 

6599 
6693 
6785 
6875 
6964 

6609 
6702 
6794 
6884 
6972 

6618 
6712 
6803 
6893 
6981 

50 

51 
52 
53 
54 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 

7185 
7267 
7348 

7024 
7110 
7193 
7275 
7356 

7033 

7118 
7202 
7284 
7364 

7042 
7126 
7210 
7292 
7372 

7050 
7135 
7218 
7300 
7380 

7059 
7143 
7226 
7308 
7388 

7067 
7152 

71':;.-, 

7316 
7396 

LOGARITHMS 


227 


N 

0 

1 

2. 

3 

4 

5 

6 

7 

8 

9 

55 

56 
57 
58 
59 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 
7731 

7435 
7513 
7589 
7664 
7738 

7443 
7520 
7597 
7672 
7745 

7451 
7528 
7604 
7679 
7752 

7459 
7536 
7612 
7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 
7627 
7701 

7774 

6O 
61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 
7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 

8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 

8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 
71 
72 

73 

74 

8451 
8513 

8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8482 
8543 
8603 
8663 
8722 

8488 
8549 
8609 
8669 

8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 
9025 

8O 
81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 
9284 

9079 
9133 
9186 
9238 
9289 

85 
86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9304 
9355 
9405 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 
9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

90 
91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 

9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

228  SECOND   COURSE    IN   ALGEBRA      [XX,  §  144 

Again,  suppose  we  wish  to  determine  log  27.6.  The  char- 
acteristic (by  §  141)  is  1.  The  mantissa,  by  Rule  I,  §  139, 
is  the  same  as  that  of  log  276 ;  and  the  latter,  as  given  in  the 
tables,  is  .4409.  Therefore,  log  27.6  =  1.4409.  Ans. 

As  a  last  example,  suppose  we  wish  to  determine  log  .0173. 
The  characteristic  (by  §  142)  is  —  2,  or  8-10.  The  mantissa, 
by  the  rule  in  §  139,  is  the  same  as  that  of  log  173  and 
the  latter,  as  obtained  from  the  tables,  is  .2380.  Therefore, 
log  .0173  =  8.2380 -10.  Ans. 

These  examples  illustrate  how  the  tables  together  with 
Rules  II  and  III,  §§  141,  142,  enable  us  to  determine  com- 
pletely the  logarithm  of  any  number  provided  it  contains 
no  more  than  three  significant  figures.  We  may  now  sum- 
marize our  results  in  the  following  rule. 

RULE  IV.  To  find  the  logarithm  of  a  number  of  three  signifi- 
cant figures  : 

1.  Look  in  the  column  headed  N  for  the  first  two  figures  of  the 
given  number.     The  mantissa  will  then  be  found  on  the  hori- 
zontal line  opposite  these  two  figures  and  in  the  column  headed 
by  the  third  figure  of  the  given  number. 

2.  Prefix  the  characteristic  according  to  Rules  II  and  III, 
§'§  141,  142. 

EXERCISES 

Determine  the  logarithm  of  each  of  the  following  numbers, 
expressing  all  negative  characteristics  as  explained  in  §  143. 

1.   451.  2.    318.  3.   861.  4.    900. 

5.  72.5      [HINT.     Note  how  log  27.6  was  obtained  in  §  144.] 

6.  7.25  7.    93.      [HINT.     Write  as  93.0] 

8.    9.      [HINT.     Write  as  9.00]  9.    .0136 

10.    .936  11.    .0036     [HINT.    Write  as  .00360] 


XX,  §  145]  LOGARITHMS  229 

12.  8560.  15.    .45  18.   .000235 

13.  .081  16.   61.7  19.   i 

14.  .8  17.   23,500.  20.  f. 

145.  To  Find  the  Logarithm  of  a  Number  of  More  Than 
Three  Significant  Figures.  Stippose  we  wish  to  determine 
log  286.7.  Here  we  have  four  significant  figures  while  our 
tables  only  tell  us  the  mantissas  of  numbers  having  three  (or 
less)  significant  figures  (as  in  §  144  and  in  the  preceding  ex- 
ercises). In  such  cases  we  proceed  as  follows : 

From  the  tables 

log     286  =  2.4564  | 

log  286.7  -  ?  Difference  =  2.4597  -  2.4564  =  .0033 

log     287  =  2.4597  J 

Since  286.7  lies  between  286  and  287,  its  logarithm  must 
lie  between  their  logarithms.  Now,  an  increase  of  one  unit 
in  the  number  (in  going  from  286  to  287)  produces  an  increase 
of  .0033  in  the  mantissa.  It  is  therefore  assumed  that  an 
increase  of  .7  in  the  number  (in  going  from  286  to  286.7)  pro- 
duces an  increase  of  .7  of  .0033,  or  .00231,  in  the  mantissa. 

Therefore  log  286.7  =  2.4564+.7  of  .0033  =  2.4564+.00231 
=  2.45871,  so  that 

log  286.7  =  2.4587  (approximately).     Ans. 

In  practice  the  answer  is  quickly  obtained  as  follows : 
The  difference  between  any  mantissa  and  the  next  higher 
one  in  the  table  (neglecting  the  decimal  point)  is  called  a 
tabular  difference.  The  tabular  difference  in  this  example  is 
4597^564,  or  33.  Taking  .7  of  this  gives  23.1,  which  (keeping 
only  the  first  two  figures)  we  call  23,  and  adding  this  to  4564 
gives  4587.  This,  therefore,  is  the  required  mantissa  of  log 
286.7,  so  that  log  286.7  =  2.4587  (approximately) .  Ans. 


230  SECOND   COURSE    IN   ALGEBRA      [XX,  §  145 

Similarly, §  in  finding  log  286.75  the  tabular  difference  (as  before) 
is  33.  Taking  .75  of  33  gives  24.75,  which  (keeping  only  two  figures) 
has  the  approximate  value  25. 

Hence  the  mantissa  of  log  286.75  is  4564+25  =4589.  Therefore 
log  286.75  =2.4589,  Ans. 

Below  are  two  examples  further  illustrating  how  the  above 
processes  are  quickly  carried  out  in  practice.  The  student 
should  form  the  habit  of  writing  the  work  in  this  form. 

EXAMPLE  1.     Determine  the  value  of  log  48. 731 

SOLUTION.     Mantissa  of  log  487  =  6875  1m,,      -, . «. 

Mantissa  of  log  488  =6884  )  Tabular  dlff^nce  =9 
.31 X9  =  2.79  =  3  (approximately). 


Hence 

mantissa  of  log  48.731  =6875+3  =  6878. 
Therefore 

log  48.731  =  1.6878     Ans. 

EXAMPLE  2.     Determine  the  value  of  log  .013403 
SOLUTION.     Mantissa  of  134  =  1271 


Mantissa  of  135  =  1303  (  Tabular  difference  =32. 

.03  X32  =  .096  =  1  (approximately). 
Hence 

mantissa  of  log  .013403  =  1271+1  =  1272. 
Therefore 

log  .013403=  -2 +.1272  =  8.1272 -10. 

Ans. 

NOTE.  The  process  which  we  have  employed  for  determining 
a  mantissa  when  it  does  not  actually  occur  in  the  tables  is  called 
interpolation.  When  examined  carefully,  it  will  be  seen  that  the  pro- 
cess is  based  upon  the  assumption  that  if  a  number  is  increased  by 
any  fractional  amount  of  itself,  the  logarithm  of  the  number  will  like- 
wise be  increased  by  the  same  fractional  amount  of  itself.  Thus,  in 
finding  the  mantissa  of  log  286.7  at  the  middle  of  p.  229,  we  assumed 
that  the  increase  of  .7  in  going  from  286  to  286.7  would  be  accom- 
panied by  like  increase  of  .7  in  the  logarithm.  Such  an  assumption, 
though  not  exactly  correct,  is  very  nearly  so  in  most  cases  and  is 
therefore  sufficiently  accurate  for  all  ordinary  purposes. 


XX,  §  146]  LOGARITHMS  231 

Tables  of  logarithms  much  more  extensive  than  those  on  pages 
226,  227  have  been  prepared  and  are  commonly  used.  See,  for  ex- 
ample, The  Macmillan  Tables.  By  means  of  these,  any  desired 
mantissa  may  usually  be  obtained  as  accurately  as  is  necessary, 
directly,  that  is  without  interpolation. 

EXERCISES 
Obtain  the  logarithm  of  each  of  the  following  n'umbers. 

1.  678.4  8.   4.806  15.  62.856 

2.  231.3  9.    1.508  16.  541.07 

3.  785.4  10.   3.276  17.  6.3478 

4.  492.6                   11.    .4567  18.  3.1416 
6.   856.8                   12.    .08346  19-  1.7096 

6.  42.17  13.   856.34  20.    .15786 

7.  9.567  14.   243.47  21.    .085679 

146.  To  Find  the  Number  Corresponding  to  a  Given  Loga- 
rithm. Thus  far  we  have  considered  how  to  determine  the 
logarithm  of  a  given  number,  but  frequently  the  problem  is 
reversed,  that  is,  it  is  the  logarithm  that  is  given  and  we  wish 
to  find  the  number  having  that  logarithm.  The  method  of 
doing  this  is  the  reverse  of  the  method  of  §§  144-145,  and  is 
illustrated  in  the  following  examples. 

EXAMPLE  1.     Find  the  number  whose  logarithm  is  1.9547 

SOLUTION.  Locate  9547  among  the  mantissas  in  the  table. 
Having  done  so,  we  find  in  the  column  N  on  the  line  with  9547  the 
figures  90.  These  form  the  first  two  figures  of  the  desired  number. 

At  the  head  of  the  column  containing  9547  is  1,  which  is  therefore 
the  third  figure  of  the  desired  number. 

Hence  the  number  sought  is  made  up  of  the  figures  901. 

The  given  characteristic  being  1,  the  number  just  found  must 
be  pointed  off  so  as  to  have  two  figures  to  the  left  of  its  decimal 
point  (Rule  II,  §  141). 

Therefore  the  number  is  90.1.     Ans. 


232  SECOND   COURSE    IN  ALGEBRA       [XX,  §  146 

EXAMPLE  2.     Find  the  number  whose  logarithm  is  0.6341 

SOLUTION.  As  in  Example  1,  we  look  among  the  mantissas  of 
the  table  to  find  6341.  In  this  case  we  do  not  find  exactly  this  man- 
tissa, but  we  see  that  the  next  less  mantissa  appearing  is  6335, 
while  the  one  next  greater  is  6345. 

The  numbers  corresponding  to  these  last  two  mantissas  are  seen 
to  be  430  and  431  respectively.  Whence,  if  x  represents  the  num- 
ber sought,  we  have 

Mantissa  of  log  430  =  6335  j  D-ff  =6  1 

Mantissa  of  log  x      =  6341  J  >  Tabular  difference  =  10. 

Mantissa  of  log  431  =6345 

Since  an  increase  of  10  in  the  mantissa  produces  an  increase  of  1  in 
the  number,  we  assume  that  an  increase  of  6  in  the  mantissa  will 
produce  an  increase  of  ^,  or  .6,  in  the  number. 

Hence  the  number  sought  has  the  figures  4306. 

Since  the  given  characteristic  is  0,  the  number  must  be  4.306 
(§141).  Ans. 

NOTE  1.  The  pupil  will  observe  that  in  Example  1  the  given 
mantissa  actually  occurs  in  the  tables,  while  in  Example  2  it  does 
not,  thus  making  it  necessary  in  this  last  case  to  interpolate.  (See 
the  Note  in  §  145.) 

NOTE  2.  The  number  whose  logarithm  is  a  given  quantity  is 
called  the  antilogarithm  of  that  quantity.  Thus  100  is  the  anti- 
logarithm  of  2,  1000  is  the  antilogarithm  of  3,  etc. 

EXERCISES 

Find  the  numbers  whose  logarithms  are  given  below. 

1.  1.8751  9.  1.4893 

2.  2.9405  10.  2.8588 

3.  0.3856  11.  3.7430 

4.  3.5866  12.  0.5240 
6.  9.6955-10  13.  0.6970 

6.  8.7152-10  14.  9.7400-10 

7.  7.4900-10  15.  8.3090-10 

8.  6.8519-10  16.  7.5308-10 


NAPIER 

(John  Napier,  1550-1617) 

Famous  as  the  inventor  of  logarithms  and  first  to  show  the  advantage  of 
using  them  in  reducing  the  labor  of  ordinary  computations.  Interested  and 
active  also  in  the  political  and  religious  controversies  of  his  day. 


XX,  §  147]  LOGARITHMS  233 

III.   THE  USE  OF  LOGAKITHMS  IN  COMPUTATION 

147.  To  Find  the  Product  of  Several  Numbers.  The  pro- 
cesses of  multiplication,  division,  raising  to  powers,  and  ex- 
traction of  roots,  as  carried  out  in  arithmetic,  may  be  greatly 
shortened  by  the  use  of  logarithms,  as  we  shall  now  show. 

Let  us  take  any  two  numbers,  for  example  25  and  37,  and 
determine  their  logarithms.     We  find  that  log  25  =  1.3979 
and  log  37  =  1.5682.     T^iis  means  (§  136)  that 
25  =  101-3979    and    37  =  101-5682 

Multiplying,  we  thus  have 

25X37  =  l01-«»»+i.B682  (§  117)  Law  j) 

The  last  equality  means  (§  136)  that 

log  (25X37)  =  1.3979+1.5682, 
or  log  (25X37)  =log  25+log  37. 

Similarly,  if  we  start  with  the  three  numbers  25, 37,  and  18 
we  can  show  that 

log  (25X37X18)=  log  25+log  37+log  18. 

Thus  we  arrive  at  the  following  important  rule. 

RULE  V.  The  logarithm  of  a  product  is  equal  to  the  sum 
of  the  logarithms  of  its  factors. 

Thus  log  (13 X. 0156X99.8)  =log  13+log  .0156+log  99.8. 

The  way  in  which  this  rule  is  used  to  find  the  value  of  the 
product  of  several  numbers  is  shown  below. 

EXAMPLE  1.     To  find  the  value  of  13 X. 0156X99.8 

SOLUTION.      log       13=   1.1139 

log  .0156=  8.1931-10 
log    99.8=   1.9991 
Adding,  11.3061 -10,  or  1.3061 

Hence,  by  Rule  V,  the  logarithm  of  the  desired  product  is  1.3061. 
It  follows  that  the  product  itself  is  the  number  whose  logarithm 
is  1.3061.  When  we  look  up  this  number  (as  in  §  146)  we  find  it  to 
be  20.23.  Hence  13  X. 0156X99.8  =20.23  (approximately).  Arcs. 


234  SECOND   COURSE    IN   ALGEBRA      [XX,  §  147 

EXAMPLE  2.     To  find  the  value  of 

8.45X.678X.0015X956X.111 

SOLUTION.      log    8.45=  0.9269 

log    .678=   9.8312-10 
log  .0015=   7.1761-10 
log     956=   2.9805 
log    .111=   9.0453-10 

/Adding,  29.9600-30=9.9600-10. 

f 

Hence,  by  Rule  V,  the  logarithm  of  the*  desired  product  is  seen  to 
be  9.9600 -10. 

Therefore  the  product  itself  is  found  (as  in  §  146)  to  be  -912 
(approximately).  Ans. 

These  examples  lead  to  the  following  rule. 
RULE  VI.     To  multiply  several  numbers  : 

1.  Add  the  logarithms  of  the  several  factors. 

2.  The  sum  thus  obtained  is  the  logarithm  of  the  product. 

3.  The  product  itself  can  then  be  determined  as  in  §  146. 

NOTE.  It  may  happen  (as  in  Example  2)  that  the  sum  of  several 
logarithms  is  negative.  In  such  cases  it  is  best  to  write  the  sum  in 
such  a  form  that  it  will  end  with  — 10,  thus  conforming  always  to 
§  143. 

EXERCISES 

Find,  by  Rule  V,  §  147,  the  value  of  each  of  the  following 
logarithms. 

1.  log  (35.1X7.29).  3.  log  (145.7 X 8.35 X. 00456). 

2.  log  (5X3.17X.0016).          4.  log  (3.456 X. 001798 XI. 456). 
Find  (by  Rule  VI,  §  147)  the  value  of 

5.   56.8X3.47 X. 735 

Check  your  answer  by  multiplying  out  the  long  way  as  in  arith- 
metic. Compare  the  two  results  and  see  how  great  was  the  error 
committed  by  following  the  short  (logarithmic)  method.  Compare 
also  the  time  required  for  the  two  methods. 


XX,  §  148]  LOGARITHMS  235 

6.  .975X42.8X3.72 

7.  896X40.8X3.75X.00489 

8.  34.56X18.16X.0157 

[HINT.     See  §  145.] 

9.  576.8X43.25X3.576X.0576 

10.  60.573X8.087X.008915X1.2387 

11.  23X23X23X23X23X23X23,  (or  237) 

12.  1.2X2.3X3.4X4.5X5.6X6.7X7.8 

13.  .31X5.198X6.831X2.584X.00312X.07568 

14.  Since  25  X  15  =  375  we  know  by  Rule  V,  §  147,  that  the 
logarithm  of  25  added  to  the  logarithm  of  15  is  equal  to  the 
logarithm  of  375.     Show  that  the  values  given  in  the  tables 
for  log  25,  log  15,  and  log  375  confirm  this  result.     Invent 
and  try  out  several  other  similar  problems  for  yourself. 

148.  To  Find  the  Quotient  of  Two  Numbers.  Let  us  take 
any  two  numbers,  for  example  41  and  29,  and  look  up  their 
logarithms.  We  find 

log  41  =  1.6128 
log  29=1.4624 
These  mean  that 

41==101.6128 

and  29  =  101-4624 

Whence,  dividing  the  first  of  these  equalities  by  the  second, 
we  obtain 

101'6128~M624     (§  117>  Law  n) 


The  last  equality  means  that 

log  (41^-29)  =  1.6128-1.4624  =  log  41-log  29. 


236  SECOND   COURSE   IN  ALGEBRA      [XX,  §  148 

This  result  illustrates  the  following  general  rule. 
RULE  VII.     The  logarithm  of  a  quotient  is  equal  to  the 
logarithm  of  the  dividend  minus  the  logarithm  of  the  divisor. 
Thus  log  (467.3  -J-.00149)  =log  467.3  -log  .00149 

The  way  in  which  this  rule  is  used  to  find  the  value  of  the 
quotient  of  two  numbers  is  shown  below. 
EXAMPLE  1.     To  find  the  value  of  236 -i-4. 15 

SOLUTION.  log  236=2.3729 

log  4.15  =0.6180 

Subtracting,  1.7549 

Hence  the  logarithm  of  the  desired  quotient  is  1.7549     (Rule  VII) 
The  number  whose  logarithm  is  1.7549  is  found  (as  in  §  146) 
to  be  56.875 

Therefore  236-^4.15=56.875  (approximately).    Ans. 

EXAMPLE  2.    To  find  the  value  of  1 . 46 -^  .00576 

SOLUTION.       log  1.46  =0.1644  =  10.1644  - 10    (See  Note  below.) 

log  .00576  =  7.7619-10 

Subtracting,  2.4025 

The  number  whose  logarithm  is  2.4025  is  found  to  be  252.64 
Therefore  1 .46  -T- .00576  =  252. 64  (approximately) .     Ans. 

Thus  we  have  the  following  rule. 

RULE  VIII.     To  find  the  quotient  of  two  numbers  : 

1.  Subtract  the  logarithm  of  the  divisor  from  the  logarithm 
of  the  dividend. 

2.  The  difference  thus  obtained  is  the  logarithm  of  the  quo- 
tient. 

3.  The  quotient  itself  can  then  be  determined  as  in  §  146. 

NOTE.  To  subtract  a  negative  logarithm  from  a  positive  one, 
or  to  subtract  a  greater  logarithm  from  a  less,  increase  the  charac- 
teristic of  the  minuend  by  10,  writing  —10  after  the  mantissa  to 
compensate.  Thus,  in  Example  2,  we  wished  to  subtract  the  nega- 
tive logarithm  7.7619-10  from  the  positive  one  0.1644.  There- 
fore 6.1644  was  written  in  the  form  10.1644-10,  after  which  the 
subtraction  was  easily  performed. 


XX,  §  149]  LOGARITHMS  237 

EXERCISES 

Find,  by  Rule  VII,  §  148,  the  value  of  each  of  the  follow- 
ing logarithms. 

1.  log(13-f-9).  3.   log  (38.76 -J-. 0017). 

2.  log  (217-5-8.16).  4.   log  (8.764 -f- 114.3). 

Find,  by  Rule  VIII,  §  148,  the  value  of  each  of  the  follow- 
ing quotients. 

5.  246 -i- 15.7 

Check  your  answer  by  dividing  out  the  long  way  as  in  arith- 
metic. Compare  the  two  results  and  see  how  great  was  the  error 
committed  by  following  the  short  (logarithmic)  method. 

6.  34.7^-5.34  8.   45.67^38.01 

7.  389.7^-4.353  9.   3.25 -f-. 00876 
[HINT.     See  §  145.]                 [HINT.     See  Note  in  §  148.] 

10.   49.6 -i- 87.3 
n    40.3X6.35 

3.72 

[HINT.     Find  the  logarithm  of  the  numerator  by  Rule  V,  §  147.] 
12    .0036X2.36  24.3  X. 695  X. 0831 

.0084  8.40  X. 216 

14.  Since  27-:-9  =  3  we  know,  by  Rule  VII,  §  147,  that  the 
logarithm  of  9  subtracted  from  the  logarithm  of  27  is  equal  to 
the  logarithm  of  3.  Show  that  the  values  given  in  the  tables 
for  log  9,  log  27,  and  log  3  confirm  this  result.  Invent  and 
try  out  several  other  similar  problems  for  yourself. 

149.  To  Raise  a  Number  to  a  Power.  Let  us  take  any 
number,  for  example  25,  and  raise  it  to  any  power,  say  the 
fourth.  We  then  have  254,  which  means  25X25X25X25. 

Hence,  by  Rule  V,  §  147,  we  must  have 
log  254  =  log  25-flog  25+log  25+log  25,  or  log  254  =  4  log  25. 


238  SECOND   COURSE   IN   ALGEBRA      [XX,  §  149 

This  illustrates  the  following  rule. 

RULE  IX.  The  logarithm  of  any  power  of  a  number  is 
equal  to  the  logarithm  of  the  number  multiplied  by  the  exponent 
indicating  the  power. 

Thus  log  3.1710  =  10  log  3.17 ;  similarly,  log.  001746  =  6  log  .00174. 

The  way  in  which  this  principle  is  used  to  raise  a  number 
to  a  power  is  shown  below. 

EXAMPLE  1.     To  find  the  value  of  2.37* 

SOLUTION.          log  2.37=  0.3747 

4 

Multiplying,  1.4988 

Hence 

Iog2.374  =   1.4988     (Rule  IX) 

The  number  whose  logarithm  is  1.4988  is  found  to  be  31.535 
Therefore 

2.374  =31.525  (approximately).     Ans. 

EXAMPLE  2.     To  find  the  value  of  .8565 
SOLUTION.  log  .856  =  9.9325  - 10 


Multiplying,  49.6625-50  =  9.6625  - 10 

The  number  whose  logarithm  is  9.6625-10  is  .4597     (§  146) 

Therefore 

.8565= . 4597  (approximately).     Ans. 

Thus  we  have  the  following  rule. 

RULE  X.     To  raise  a  number  to  a  power  : 

1.  Multiply  the  logarithm  of  the  number  by  the  exponent 
indicating  the  power. 

2.  The  result  thus  obtained  is  the  logarithm  of  the  answer. 

3.  The  answer  itself  can  then  be  determined  as  in  §  146. 

EXERCISES 

Find,  by  Rule  IX,  §  149,  the  value  of  each  of  the  following 
logarithms. 

1.   log  166      2.   log  3.123      3.   log  .01762       4.   log  36.644 


XX,  §  150]  LOGARITHMS  239 

Find,  by  Rule  X,  §  149,  the  value  of  each  of  the  following 
expressions. 

5.  8.823 

Check  your  answer  by  raising  8.82  to  the  third  power  as  in 
arithmetic.  Compare  the  two  results  and  see  how  great  was  the 
error  committed  by  following  the  short  (logarithmic)  method. 

6.  4'.124  7.   4.1234 

8.  .1755     [HINT.     See  Ex.  2  in  §  149.] 

9.  813X.0152    [HINT.     Combine  the  rules  of  §§  147  and  149.] 
10.   43X8.92X.0753 

8.76X53.9X4.53 
'   2.32X3.15X5.143 
[HINT.     Use  Rules  VI,  VIII,  X.] 

12.  Since  93  =  729  we  know,  by  Rule  IX,  §  149,  that  three 
times  the  logarithm  of  9  is  equal  to  the  logarithm  of  729. 
Show  that  the  values  given  in  the  tables  for  log  9  and  log  729 
confirm  this  result.  Invent  and  try  out  several  other  similar 
problems  for  yourself. 

150.  To  Extract  Any  Root  of  a  Number.  Let  us  take  any 
number,  for  example  36,  and  consider  any  root  of  it,  say  the 
fifth,  that  is,  let  us  consider  \/36l 

Supposing  x  to  be  the  value  of  the  desired  root,  we  have 
z5  =  36.  (§  118) 

Now  the  logarithm  of  the  first  member  of  this  equality  is 
equal  to  5  log  x  by  Rule  IX. 

Hence  5  log  x  =  \og  36,  or  log  x  =  ^  log  36. 

This  illustrates  the  following  rule. 

RULE  XI.  The  logarithm  of  the  root  of  a  number  is  equal 
to  the  logarithm  of  the  radicand  divided  by  the  index  of  the  root. 

Thus  log  ^73=^  log  2.73 ;  similarly,  log  -v/.01685=|  log  .01685. 

The  way  in  which  this  principle  is  used  to  extract  the  roots 
of  numbers  in  arithmetic  will  now  be  shown. 


240  SECOND   COURSE    IN   ALGEBRA      [XX,  §  150 

EXAMPLE  1.    To  find  the  value  of  'V/85.2 
SOLUTION.        log  85.2  =  1  .9304, 

so  that  -J-  of  log  85.2=0.4826. 

Hence  log  ^85^2"  =0.4826.     (Rule  XI) 

The  number  whose  logarithm  is  0.4826  is  3.038      (§  146) 
Therefore  ^85.2=3.038  (approximately).     Ans. 

EXAMPLE  2.    To  find  the  value  of  v^.0875 
SOLUTION.      log  .0875  =  8.9420  -  10, 

so  that         i  of  log  .0875  =  ^8.9420  -  10)  =  i(48.9420  -50) 
=  9.7884  -  10.     (See  Note  below.) 

The  number  whose  logarithm  is  9.7884-10  is  .6143     (§  146) 
Therefore  ^.0875  =  .  6143  (approximately).     Ans. 

These  examples  lead  to  the  following  rule. 
RULE  XII.     To  find  any  root  of  any  number. 

1.  Divide  the  logarithm  of  the  number  by  the  index  of  the  root. 

2.  The  quotient  thus  obtained  is  the  logarithm  of  the  desired 
root. 

3.  The  root  itself  can  then  be  determined  as  in  §  146. 
NOTE.     To  divide  a  negative  logarithm,  write  it  in  a  form  where 

the  negative  part  of  the  characteristic  may  be  divided  exactly  by 
the  divisor  giving  —  10  as  quotient.  Thus,  in  Example  2,  we  wrote 
8.9420-10  in  the  form  48.9420-50  after  which  the  division  by  5 
was  easily  done  and  resulted  in  a  form  ending  in  —  10. 

EXERCISES 

Find,  by  Rule  XI,  §  150,  the  value  of  each  of  the  following 
logarithms. 

1.  log  vT6  2.  log  v^332  3.  log  ^0175  4.  log  v/38^6" 
Find,  by  Rule  XII,  §  150,  the  value  of  each  of  the  following. 
5. 


Check  your  answer  by  extracting  the  square  root  of  315  (correct 
to  three  decimal  places)  as  in  arithmetic.  Compare  the  two  results 
and  see  how  great  was  the  error  committed  by  following  the  short 
(logarithmic)  method. 


XX,  §  150]  LOGARITHMS  241 


8. 

[HINT.     See  Example  2  in  §  150.] 

9.    -^8.76  X.  0153 

[HINT.     Use  Rules  IX  and  XI.]          _ 

1576  X  9.  132 


11. 


3.8X5.323 

12.  Since  A/49  =  7  we  know,  by  Rule  XI,  §  150,  that  one 
half  the  logarithm  of  49  is  equal  to  the  logarithm  of  7.  Show 
that  the  values  given  in  the  tables  for  log  49  and  log  7  con- 
firm this  result.  Invent  and  try  out  several  other  similar 
problems  for  yourself. 

APPLIED    PROBLEMS 

Solve  the  following  exercises  by  logarithms. 

1.  How  many  cubic  feet  of  air  are  there  in  a  schoolroom 
whose  dimensions  are  50.5  ft.  by  25.3  ft.  by  10.4  ft.  ? 

2.  How  many  gallons  will  a  rectangular  tank  hold  whose 
dimensions  are  8  ft.  10  in.  by  9  ft.  3  in.  by  10  ft  1  in.  ? 

3.  How  much  wheat  will  a  cylindrical  bin  hold  if  the 
diameter  of  the  base  is  9  ft.  5  in.  and  the  height  is  40  ft.  4  in.  ? 

4.  How  much  would  a  sphere  of  solid  cork  weigh  if  its 
diameter  was  4  ft.  3  in.,  it  being  known  that  the  specific 
gravity  of  cork  is  .24?     (See  Example  14  (e),  page  6.) 

[HINT.  To  say  that  the  specific  gravity  of  cork  is  .24  means  that 
any  volume  of  cork  weighs  .24  times  as  much  as  an  equal  volume  of 
water.  Water  weighs  62.5  pounds  per  cubic  foot.] 

5.  The  diameter  d  in  inches  of  a  wrought-iron  shaft  re- 
quired to  transmit  h  horse  power  at  a  speed  of  n  revolutions 

per  minute  is  given  by  the  formula  d  =  \ Find  the 

*    n 

diameter  required  when  135  horse  power  is  to  be  transmitted 
at  a  speed  of  130  revolutions  per  minute. 


242  SECOND   COURSE    IN   ALGEBRA      [XX,  §  150 

6.   The  amount  to  which  P  dollars  will   accumulate  at 
r%  compound  interest  in  n  years  is  given  by  the  formula 


Find  A  if  P  =  $500,  r  =  5,  and  n  =  W. 
Find  A  if  P  =  $100,  r  =  3.5,  and  n=15. 

7.  By  means  of  Formula  3  of  §  65,  find  the  area  of  the 
triangle  whose  sides  are  3.15  in.,  4.87  in.,  and  2.68  in. 

8.  The  height  H  of  a  mountain  in  feet  is  given  by  the 
formula 

H  =  49,00(/^— rVl +—\ 
\R+rJ\       900  / 

where  R,  r  are  the  observed  heights  of  the  barometer  in  inches 
at  the  foot  and  at  the  summit  of  the  mountain,  and  where  T, 
t  are  the  observed  Fahrenheit  temperatures  at  the  foot  and 
summit. 

Find  the  height  of  a  mountain  if  the  height  of  the  barom- 
eter at  the  foot  is  29.6  inches  and  at  the  summit  25.35 
inches,  while  the  temperature  at  the  foot  is  67°  and  at  the 
summit  32°. 

9.  By  means  of  the  formula  in  Ex.  6  answer  the  follow- 
ing question :     How  long  will  it  take  a  sum  of  money  to 
double  itself  if  placed  at  compound  interest  at  5  %  ? 

14.2  years.     Ans. 

GENERAL  LOGARITHMS 

*161.  Logarithms  to  Any  Base.  In  §  136  we  defined  the  loga- 
rithm of  a  number  as  the  power  to  which  10  must  be  raised  to  obtain 
that  number.  Thus,  from  such  equalities  as  102  =  100,  103  =  1000, 
etc.,  we  had  log  100  =  2,  log  1000  =  3,  etc.  Strictly  speaking,  this 
defines  the  logarithm  of  a  number  to  the  base  10,  or,  as  it  is  usually 
called,  a  common  logarithm. 

We  may  and  frequently  do  use  some  other  base  than  10.  For 
example,  since  32=9,  33=27,  34=81,  etc.,  we  can  say  that  the 
logarithm  of  9  to  the  base  3  is  2,  the  logarithm  of  27  to  the  base  8  is  3, 


XX,  §  154]  LOGARITHMS  243 

the  logarithm  of  81  to  the  base  3  is  4,  etc.  The  usual  way  of  denot- 
ing this  is  to  write  Iog39  =2,  Iog327  =3,  Iog381  =4,  etc.  Observe  that 
the  number  being  used  as  the  base  is  thus  placed  to  the  right  and 
just  below  the  symbol  log. 

Similarly,  we  have  Iog216=4,  Iog864  =  2,  Iog6125=3,  etc. 

Thus  we  have  the  following  general  definition.  The  logarithm  of 
any  number  x  to  a  given  base  a  is  the  power  of  a  required  to  give  x.  It  is 
written  \ogax.  Any  positive  number  except  1  may  be  used  as  the  base. 

NOTE.  When  the  base  a  is  taken  equal  to  10  (that  is,  in  the  usual 
case)  we  write  simply  log  x  instead  of 


EXERCISES 

State  first  the  meaning  and  then  the  value  of 
1.   Iog24.  2.   Iog28.  3.   Iog416.  4.   logsf 

6.   Iog2i.  6.   log^.  7.   Iog5.2  8.   Iog832. 

*152.  Logarithm  of  a  Product.  We  can  now  show  that  Rule  V, 
§  147,  holds  true  whatever  the  base.  That  is,  if  M  and  N  are  any  two 
numbers,  and  a  the  base,  then 

logaMN=logaM+logaN. 

PROOF.  Let  x  =  logaM  and  y  =  logaN.  Then  ax  =  M  and  a"  =  N 
(§151).  Hence  a*-a«  =  MN,  or  ax+«  =  MN  (§117,  Law  I). 
But  the  last  equality  means  that 

logaMN=x+y=\ogaM+logaN.  (§  151) 

*153.  Logarithm  of  a  Quotient.  Rule  VII,  §  148,  holds  true 
whatever  the  base.  That  is,  if  M  and  N  are  any  two  numbers,  then 

lQg.(M+N)  =  \OgaM-logaN. 

PROOF.  Let  x  =  log0M  and  y  =  \ogaN.  Then  ax  =  M  and  a«  =  N. 
(§  151).  Hence,  a*  -=-  a"  =  M  -=-  N,  OTax~v  =  M  +  N  (§  117,  Law  II). 
But  the  last  equality  means  that 

loga(M  +  N)=x-y=logaM-logaN.  (§  151) 

*164.  Logarithm  of  a  Power  of  a  Number.  Rule  IX,  §  149,  holds 
true  whatever  the  base.  That  is,  if  M  is  any  number  and  n  any  (posi- 
tive integral)  power,  then 

log0Mre  =  n  logaM. 

PROOF.  Letz  =  logaAf.  Then  a*  =  M  (§  151)  and  hence  anx  =  Mn 
(§  117,  Law  III).  But  the  last  equality  means  that 

log0Mn  =  nx=n  log0M.  (  §  151  ) 


244  SECOND   COURSE   IN  ALGEBRA      [XX,  §  155 

*155.  Logarithm  of  a  Root  of  a  Number.  Rule  XI,  |  150,  holds 
true  whatever  the  base.  That  is,  if  M  is  any  number  and  n  any  (posi- 
tive integral)  root,  then 


=     log0M. 
n 

PROOF.  Let  x=logaM  Then  ax  =  M  (§151)  and  hence 
(0*)i/»  =  M1/n,  ora*/n  =  VM"  (§  121).  But  the  last  equality  means 
that 


n     n 

*166.  Summary.  From  the  results  established  in  §§  151-155  it 
appears  that  Rules  V-XII,  §§  147-150,  are  not  only  true  when  the 
base  is  10  (as  was  there  taken)  but  they  are  true  for  any  base. 
Complete  tables  have  been  worked  out  for  various  bases  other  than 
10,  but  we  shall  not  consider  them  further  here. 

NOTE.  The  reason  why  1  cannot  be  used  as  a  base  is  that  1  to 
any  power  is  equal  to  1,  that  is,  we  cannot  get  different  numbers  by 
raising  1  to  different  powers. 

*157.  Historical  Note.  Logarithms  were  first  introduced  and 
employed  for  shortening  computation  by  JOHN  NAPIER  (1550-1617), 
a  Scotchman.  (See  the  picture  facing  p.  233.)  However,  he  did 
not  use  the  base  10,  this  being  first  done  by  the  English  mathemati- 
cian BRIGGS  (1556-1631),  who  computed  the  first  table  of  common 
logarithms  and  did  much  to  bring  logarithms  into  general  use. 

*158.  Calculating  Machines.  The  Slide-Rule.  Machines  have 
been  invented  and  are  now  coming  into  very  general  use,  especially 
by  engineers,  by  which  the  processes  of  multiplication,  division, 
involution,  and  evolution  can  be  immediately  performed.  The 
construction  of  these  machines  depends  upon  the  principles  of 
logarithms,  but  to  describe  the  machines  and  their  methods  of 
working  would  take  us  beyond  the  scope  of  this  text.  The  simplest 
machine  of  this  kind  is  the  slide  rule,  the  use  of  which  is  easily 
understood.  A  simple  slide  rule  with  directions  is  inexpensive 
and  may  ordinarily  be  secured  from  booksellers. 


IA 

A 

I   c      (( 

l,I|p!.l  ^111! 

i""l"  ^     !    ^ 

D 

FIG.  70.    THE  SLIDE  RULE. 


PAET  III.     SUPPLEMENTARY  TOPICS 

CHAPTER  XXI 
FUNCTIONS 

159.  The  Function  Idea.  In  ordinary  speech  we  make 
such  statements  as  the  folio  whig : 

1.  The  area  of  a  circle  depends  upon  the  length  of  its 
radius. 

2.  The  time  it  takes  to  go  from  one  place  to  another  de- 
pends upon  the  distance  between  them. 

3.  The  power  which  an  engine  can  exert  depends  upon  the 
pressure  per  square  inch  of  the  steam  in  the  boiler. 

Another  way  of  stating  these  facts  is  as  follows : 

1.  The  area  of  a  circle  is  a  function  of  the  length  of  its 
radius. 

2.  The  time  it  takes  to  go  from  one  place  to  another  is  a 
function  of  the  distance  between  them. 

3.  The  power  which  an  engine  can  exert  is  a  function  of 
the  pressure  per  square  inch  of  the  steam  in  the  boiler. 

The  idea  thus  conveyed  by  the  word  function  is  that  we 
have  one  magnitude  whose  value  is  determined  as  soon  as  we 
know  the  value  of  some  other  one  (or  more)  magnitudes  upon 
which  the  first  one  depends.  This  idea  is  at  once  seen  to  be 
universal  in  everyday  experience  and  for  that  reason  it  be- 
comes of  great  importance  in  mathematics,  f  In  the  present 

t  The  extended  formal  study  of  the  function  idea  enters  into 
that  branch  of  mathematics  known  as  the  Calculus. 

245 


246  SECOND   COURSE    IN   ALGEBRA    [XXI,  §  159 

chapter  we  shall  indicate  briefly  how  it  is  related  to  some  of 
the  subjects  treated  in  the  preceding  chapters,  noting  es- 
pecially the  significance  of  the  idea  when  considered  graphi- 
cally. 

160.  Types  of  Algebraic  Functions.  An  expression  of  the 
form 

(1)  «wr+fli, 

where  the  coefficients  aQ  and  ai  have  any  given  values  (except 
a0  must  not  be  0)  is  called  a  linear  function  of  x.  Observe 
that  every  such  expression  depends  for  its  value  upon  the 
value  assigned  to  #,  and  is  determined  as  soon  as  x  is  known. 
Hence  it  is  a  function  of  x  in  the  sense  explained  in  §  159. 
It  is  called  a  linear  function  since  it  is  of  the  first  degree  in  x. 
(Compare  §  26.) 

For  example,  2  x  +3  is  a  linear  function  of  x.  Here  we  have  the 
form  (1)  inwhicha0  =  2  and  ai=3.  Similarly,  3x—  2,  x—  4,-z+i 
and  3x  are  linear  functions  of  x.  (Why?) 

Likewise,  3^+2  is  a  linear  function  of  t,  while  —  r+5  is  a  linear 
function  of  r,  etc. 

As  an  example  of  a  linear  function  in  everyday  experience,  sup- 
pose that  in  Fig.  71  a  person  starts  from  the  point  P  and  moves  to 
the  right  at  the  rate  of  15  miles  per  hour,  and  let  Q  be  the  point  10 


S E 

h— 10— >! 


FIG.  71. 

miles  to  the  left  of  P.  Then  we  may  say  that  the  distance  of  the 
traveler  from  Q  is  a  linear  function  of  the  time  he  has  been  traveling, 
for  if  t  represent  the  number  of  hours  he  has  been  traveling,  his  dis- 
tance from  P  is  15 1  (see  §  62)  and  hence  his  distance  from  Q  is 
15  £  +  10.  This  is  seen  to  be  a  linear  function  of  t,  being  of  the  form 
(1)  in  which  a0  =  15  and  01  =  10. 

.   Likewise,  the  interest  which  a  given  principal,  P,  will  yield  in  one 
year  is  a  linear  function  of  the  rate,  for,  if  r  be  the  rate,  the  interest 


XXI,  §  160]  FUNCTIONS  247 

in  question  is  given  by  the  formula  P  X-^:,or— —  r,  and  this  is  seen 

,  -LvJO        J.LHJ 

p 

to  be  of  the  form  (1)  in  which  «o  =  rrr:,  and  ai=0. 

J_  \j\J      .___  - 

An  expression  of  the  form 

(2)  a0x2+aix+a2, 

where  the  coefficients  a0,  «i,  and  a2  have  any  given  values 
(except  that  a0  must  not  be  0)  is  called  a  quadratic  function 
of  x. 

For  example,  2rr2+3  x  —  1  is  a  quadratic  function  of  x  because  it 
is  of  the  form  (2)  in  which  a0  =  2,  a\  =  3,  02  =  —  1.  Likewise,  x2  +-|-  x  ; 
x2+i;  —  z2+3  x  ;  5  x2 ;  z2  are  quadratic  functions  of  x.  (Why?) 

Again,  we  may  say  that  the  area  of  a  square  is  a  quadratic  func- 
tion of  the  length  of  one  side,  for  if  x  be  the  length  of  side,  the  area 
is  xz  and  this  is  of  the  form  (2)  in  which  a0  =  1,  a\  =  a2  =0. 

Similarly,  the  area  of  a  circle  is  a  quadratic  function  of  the 
radius.  (Why?) 

An  expression  of  the  form 

(3)  flo*3+ai*2+a2Jt+a3, 

where  the  coefficients  a0,  «i,  «2  and  a3  have  any  given  values 
(except  that  a0  must  not  be  0)  is  called  a  cubic  function  of  x. 

For  example,  3  x3-x*+±x-l ;   4z3-z;   z3-2x2  +  l;   5x3;   x3, 
etc.     (Why?) 

Again,  we  may  say  that  the  volume  of  a  cube  is  a  cubic  function 
of  the  length  of  one  edge.  (Why?)  !  Also,  the  volume  of  a  sphere 
is  a  cubic  function  of  the  radius.  (Why?) 

It  may  now  be  observed  that  the  expressions  (1),  (2),  and 

(3)  are  but  special  forms  of  the  more  general  expression 

(4)  a0xn+ai^n~1+fl2Xn-2H \-an_iX+an 

where  it  is  understood  that  n  can  be  any  positive  integer, 
while  the  coefficients  OQ,  ai,  a2,  •••  a,,  have  any  given  values 
(except  that  a0  must  not  be  0).  This  is  called  the  general 


248  SECOND   COURSE   IN  ALGEBRA     [XXI,  §  160 

integral  rational  function  of  x,  or,  more  simply,  a  polynomial 
in  x.     It  reduces  to   the   linear  function  (1)  when  n=l; 
to  the  quadratic  function  (2)  when  n  =  2  ;   etc. 
Expressions  such  as 

Vx, 


and  all  others  composed  merely  of  powers  or  roots  (or  both) 
of  x  are  classed  under  the  name  of  algebraic  functions. 
Since  all  functions  of  the  form  (4)  are  composed  of  integral 
powers  only  of  x,  they  are  but  special  cases  of  the  algebraic 
functions  just  mentioned. 

EXERCISES 

1.  Show  that  the  thickness  of  a  book  is  a  linear  function 
of  the  number  of  its  pages. 

[HINT.  Let  x  be  the  number  of  pages,  d  be  the  thickness  of  each 
page,  and  D  the  thickness  of  each  cover.  Now  build  up  the  formula 
for  the  thickness  of  the  book  and  note  which  of  the  functional 
types  in  §  160  is  present.] 

2.  The  supply  of  gasoline  in  a  tank  was  very  low,  its 
depth  being  but  1  inch  all  over  the  bottom,  when  it  was  re- 
plenished from  a  pipe  which  delivered  3  gallons  per  minute. 
Show  that  the  amount  in  the  tank  at  any  moment  during  the 
filling  was  a  linear  function  of  the  time  since  the  filling  began. 

3.  Show  that  the  force  which  a  steam  engine  has  at  any 
moment  at  its  cylinder  is  a  linear  function  of  the  area  of  the 
piston  ;  also  that  it  is  a  linear  function  of  the  boiler  pressure 
of  the  steam  per  square  inch. 

4.  A  certain  room  contains  a  number  of  16-candle-power 
electric  lights  and  a  number  of  Welsbach  gas-burners.     Show 
that  the  amount  of  illumination  at  any  time  is  a  linear 
function  of  the  number  of  electric  lights  turned  on.     Is  this 
true  regardless  of  the  number  of  gas-burners  already  lighted  ? 


XXI,  §  160]  FUNCTIONS  249 

5.  Show  that  the  perimeter  of  a  square  is  a  linear  func- 
tion of  the  length  of  one  side ;    also  that  the  circumference 
of  a  circle  is  a  linear  function  of  its  radius. 

6.  Show  that  if  each  side  of  a  square  be  increased  by  x, 
the  corresponding  increase  in  the  area  will  be  a  quadratic 
function  of  x. 

[HINT.  Let  a  =  the  length  of  one  side  of  the  original  square. 
Then  the  area  is  a2  and  the  area  of  the  new  square  is  (a  +z)2.  Now 
formulate  the  expression  for  the  increase  in  area.] 

7.  Show  that  if  the  radius  of  a  circle  be  increased  by  x, 
the  corresponding  increase  in  area  will  be  a  quadratic  func- 
tion of  x. 

8.  Show  that  if  the  edge  of  a  cube  be  increased  by  x 
the  corresponding  increase  in  volume  will  be  a  cubic  function 
of  x.    State  and  prove  the  corresponding  statement  for  a 
sphere. 

9.  Show  that  if  y  varies  directly  as  x  (see  §  113),  then  y 
is  a  linear  function  of  x.     Is  the  converse  of  this  statement 
necessarily  true,  namely  if  y  is  a  linear  function  of  x,  then  y 
varies  directly  as  x  ? 

10.  When  y  varies  as  the  square  of  x,  to  which  one  of  the 
functional  types  mentioned  in  §  160  does  y  belong?     Answer 
the  same  question  when  y  varies  inversely  as  x ;  when  y  varies 
inversely  as  the  square  of  x. 

11.  A  certain  linear  function  of  x  takes  the  value  5  when 
x  =  l  and  takes  the  value  8  when  x  =  2.     Determine  com- 
pletely the  form  of  the  function. 

SOLUTION.  Since  the  function  is  linear,  it  is  of  the  form  aox  +«i. 
Since  this  expression  must  (by  hypothesis)  be  equal  to  5  when  x  =  1, 
we  have  a0  •  l+ai=5.  Likewise,  placing  x=2,  gives  a0  •  2+ai=8. 
Solving  these  two  equations  for  a0  and  ai  we  obtain  oo=3,  ai=2. 
The  desired  function  is  therefore  3  x  +2.  Ans. 


250 


SECOND   COURSE    IN   ALGEBRA    [XXI,  §  160 


12.  A  certain  linear  function  of  x  takes  the  value   14 
when  x  =  3,  and  takes  the  value  —6  when  x=  —  1.     Deter- 
mine completely  the  form  of  the  function. 

13.  A  certain  quadratic  function  takes  the  value  0  when 
x  =  1,  and  the  value  1  when  x  =  2,  and  the  value  4  when  x  =  3. 
Determine  completely  the  form  of  the  function. 

14.  Show  that  the  area  of  any  triangle  is  an  algebraic 
function  of  the  sum  of  its  three  sides.     (See  Formula  3  in 
§65.) 

161.  Functions  Considered  Graphically.  By  the  graph 
of  a  function  is  meant  the  line  or  curve  which  results  when 
some  letter,  as  y,  is  placed  equal  to  the  function  and  the  graph 
is  drawn  of  the  equation  thus  obtained.  The  purpose  of 
the  graph  is  to  bring  out  clearly  and  quickly  to  the  eye  the 
relation  between  the  given  function  and  the  quantity  (vari- 
able) upon  which  it  depends  for  its  values. 

The  method  of  drawing  such  graphs  is  precisely  the  same 
as  that  given  in  §  29,  p.  43  for  equations  of  the  first  degree, 
and  in  §  57,  p.  90,  for  quadratic  equations. 

Thus,  in  order  to  obtain  the  graph  of  the  function  x*,  we  place 
y=*xz  and  proceed  to  draw  the  graph  of  this  equation  in  the  way 
explained  in  §  29,  that  is,  we  assign  various  values  to  x  and  compute 
(from  this  equation)  the  corresponding  values  of  y,  then  we  plot 
each  point  thus  obtained  and  finally  draw  the  smooth  curve  passing 
through  all  such  points. 

Below  is  a  table  of  several  values  of  x  and  y  thus  computed; 
and  the  graph  is  shown  in  Fig.  72. 


When  x  = 

-2 

-1 

0 

1 

2 

3 

4 

then     y  = 

-8 

-1 

0 

1 

8 

27 

64 

The  portion  of  the  curve  lying  to  the  right  of  the  y-a,xis  extends  up- 
ward indefinitely,  while  the  portion  to  the  left  of  the  same  axis  ex- 
tends downward  indefinitely.  Note  that,  from  the  way  this  curve  has 


XXI,  §  161] 


FUNCTIONS 


251 


been  drawn,  it  at  once  brings  out  to  the  eye 
the  value  of  the  given  function  x3  for  any 
value  of  the  letter  x  upon  which  this  func- 
tion depends,  the  function  values  being  the 
ordinates  (§  28)  of  the  points  on  the  curve. 
For  example,  at  x  =  2  the  corresponding 
ordinate  measures  8,  which  is  the  function 
value  then  present. 

This  curve  may  be  used  as  a  graphical 
table  of  cubes  of  numbers.  Thus,  if  x  =  1.5, 
2/  =  3.4,  approximately,  etc.  Likewise,  if  y 
is  given  first,  the  curve  shows  the  cube  root 
of  y;  for  example,  if  y=4,  x  is  about  1.6. 
The  figure  may  be  drawn  by  the  student  on 
a  much  larger  scale ;  the  values  of  x  and  y 
can  be  read  much  more  accurately  from 
such  a  figure  than  from  the  small  figure  on  , 
this  page. 

Another  means  of  improving  the  accu- 
racy of  the  figure  is  to  take  a  longer  dis- 
tance on  the  horizontal  line  to  represent  one 
unit  than  is  taken  to  represent  one  unit  on 
the  vertical  scale. 


FIG.  72. 


The  graph  of  every  linear  function  is  a  straight  line.     The 
graph  of  every  other  algebraic  function  is  a  curved  line. 


-B 


FIG.  73. 


For  example,  in  considering  the  graph 
of  the  linear  function  \x— 5,  we  place 
y=\ x— 5.  But  this  is  an  equation  of  the 
first  degree  between  x  and  y  and  hence 
(§  29)  its  graph  is  a  straight  line.  Fig.  73 
shows  the  result. 

Note  that  the  graph  cuts  the  z-axis 
in  one  point.  The  abscissa  of  this  par- 
ticular point  is  4,  which  indicates  that  4 
is  the  root,  or  solution,  of  the  equation 
f  x— 5=0,  for  it  is  this  value  of  x  that 
makes  y=Q. 


252 


SECOND   COURSE    IN   ALGEBRA     [XXI,  §  161 


The  graph  of  every  quadratic  function  belongs  to  the  class 
of  curves  known  as  parabolas.  A  parabola  resembles  in  form 
an  oval,  open  at  one  end.  It  never  cuts  the  z-axis  in  more 
than  two  points. 

Fig.  74  shows  the  graph  of  the  quadratic  function  xz+x-2. 
Note  that  the  curve  cuts  the  z-axis  at  two  points  whose  abscis- 
sas are  —  2  and  1,  respectively.  This  indicates  that  —  2  and  1  are 
the  roots  of  the  quadratic  equation  x2+x  —2=0. 


1-0 


FIG.  74. 


FIG.  75. 


The  general  form  of  the  graph  of  a  cubic  function  is  that 
of  an  indefinitely  long  smooth  curve  which  cuts  the  x-axis 
in  no  more  than  three  points. 

Fig.  75  shows  the  graph  of  the  cubic  function  x3  —  3  z2  —  x  +3.  It 
cuts  the  rr-axis  at  three  points  whose  abscissas  are  respectively  —1, 
1,  and  3.  These  values,  therefore,  are  the  roots  of  the  cubic  equa- 
tion z8-3z2-z+3=0. 


XXI,  §  161] 


FUNCTIONS 


253 


Similarly,  the  general  form  of  the  graph  of  the  rational 
integral  function  of  the  fourth  degree  is  that  of  an  indefinitely 
long  smooth  curve  which  cuts  the  x-axis  in  no  more  than 
four  points.  And  it  may  be  said  likewise  that  the  graph 
of  the  general  integral  function  of  degree  n  (see  (4),  §  160) 
is  an  indefinitely  long  smooth  curve  which  cuts  the  z-axis 
in  no  more  than  n  points. 

Fig.  76  shows,  for  example,  the  graph  of  2  x4  —5  x3  +5  x  -  2,  this 
being  a  function  of  the  fourth  degree.  The  four  points  where  the 
curve  cuts  the  z-axis  have  abscissas  which  are  equal  respectively 
to  —1,  -1-,  1,  and  2.  These  values,  therefore,  are  the  roots  of  the 
equation  2  x4  —  5  z3+5  x  —  2  =  0. 


FIG.  76. 


FIG.  77. 


Fractional  expressions  give  rise  to  more  complex  graphs,  which 
may  have  more  than  one  piece.  Fig.  77  shows,  for  example,  the 
graph  of  1/rc.  If  we  let  y  =  l/x,  y  varies  inversely  as  x  (§  110). 
The -curve  is  therefore  similar  to  those  drawn  in  §  115,  Fig.  69. 
The  graph  consists  of  two  branches  and  belongs  to  the  class  of 
curves  known  as  hyperbolas.  These  we  have  already  met  in  §  78. 


254  SECOND   COURSE    IN   ALGEBRA     [XXI,  §  161 

EXERCISES 

Draw  the  graphs  of  the  following  functions  by  plotting 
several  points  on  each  and  drawing  the  curve  through  them. 
Try  to  plot  enough  points  so  that  the  form  and  location  of 
the  various  waves,  or  arches,  of  the  curve  will  be  brought 
out  clearly,  as  in  the  figures  of  §  161.  Note  how  many 
times  the  curve  cuts  the  x-axis  and  make  such  inferences  as 
you  can  regarding  the  roots  of  the  corresponding  equation. 

[HINT.  When  the  graph  of  a  quadratic  function  fails  to  cut  the 
x-axis,  this  indicates  that  the  roots  of  the  corresponding  quadratic 
equation  are  imaginary.  (See  §§  57,  60.)  Similarly,  when  the 
graph  of  a  cubic  function  cuts  the  z-axis  in  but  one  point,  this  indi- 
cates that  there  is  but  one  real  root  to  the  corresponding  equation, 
the  other  two  roots  being  imaginary.  In  general,  the  number  of 
times  the  graph  cuts  the  x-axis  indicates  the  number  of  real  roots 
of  the  corresponding  equation,  the  number  of  imaginary  roots  being 
the  degree  of  the  equation  minus  the  number  of  real  roots.] 

1.   3x+4.          2.   x.  3.  xz-x-2.  4.   z2-4. 

5.   x2+l.          6.   x*-3x*-x+3.          7. 


CHAPTER  XXII 
MATHEMATICAL   INDUCTION  —  BINOMIAL   THEOREM 

162.    Mathematical    Induction.       The    three    following 
purely  arithmetic  relations  are  easily  seen  to  be  true  : 


l+2+3  = 


We  might  at  once  infer  from  these  that  if  n  be  any  positive 
integer,  there  exists  the  algebraic  relation 

(1)  1+2+3+4+  •••+n  =  £(n+l), 

« 

the  dots  indicating  that  the  addition  of  the  terms  on  the  right 
continues  up  to  and  including  the  number  n. 
For  example,  if  n  =  8,  this  would  mean  that 

l+2+3+4+5+6+7+8=f(8  +  l).  - 
Again,  if  n  =  10,  it  would  mean  that 

1+2+3+4+5+6+7+8+9  +  10=^(10  +  1). 

That  these  are  indeed  true  relations  is  discovered  as  soon  as  we 
simplify  them.  Let  the  pupil  convince  himself  on  this  point. 

It  is  now  to  be  carefully  observed  that  the  inference  just 
made,  namely  that  (1)  is  true-  for  any  n,  is  not  yet  justified, 
strictly  speaking,  from  anything  we  have  done,  for  we  have 
only  shown  that  (1)  holds  good  for  certain  special  values  of  n, 
and  we  could  never  hope  to  do  more  than  this  however  long 
we  continued  to  try  out  the  formula  in  this  way. 

Something  more  than  a  knowledge  of  special  cases  must  always 
be  known  before  any  perfectly  certain  general  inference  can  be  made. 
For  example,  the  fact  that  Saturday  was  cloudy  for  38  weeks  in  suc- 
cession gives  no  certain  information  that  it  will  be  so  on  the  39th 
week. 

255 


256  SECOND   COURSE    IN  ALGEBRA     [XXII,  §  162 

We  shall  now  show  how  the  general  formula  (1)  may  be 
established  free  from  all  objection,  that  is  in  a  way  that 
leaves  no  possible  question  as  to  its  truth  in  all  cases. 

Let  r  represent  any  one  of  the  special  values  of  n  for  which 
we  know  (1)  to  be  true.  Then 

(2)  1+2+3+4+  ."+r=|(r+l). 

Let  us  add  (r+1)  to  both  sides.     The  result  is 

1+2+3+4+.-  H-r+(r+l)  =  £(r+l)  +  (r+l). 

2i 

In  the  second  member  of  the  last  equation  we  may  write 


while  the  first  member  has  the  same  meaning  as 


Thus,  (2)  being  given  us,  it  follows  that  we  may  write 
(3) 


But  (3)  is  seen  to  be  precisely  the  same  as  (2)  except  that 
r+1  now  replaces  r  throughout.  Stated  in  words,  this  re- 
sult means  that  if  (1)  is  true  when  n  =  r,  as  we  have  supposed, 
then  it  holds  true  necessarily  for  the  next  greater  value  of  n, 
which  is  r+1. 

The  original  fact  which  we  wished  to  establish  (namely, 
that  (1)  is  true  for  any  n)  now  follows  without  difficulty. 
In  fact,  we  know  (see  beginning  of  this  section)  that  (1)  is 
true  when  n  =  4,  from  which  it  now  follows  that  it  must  be 
true  also  when  n  =  5.  Being  true  when  n  =  5,  the  same 
reasoning  says  it  must  be  true  also  when  n  =  6.  Being  true 
when  n  =  6,  it  must  be  likewise  true  when  n  =  7.  Proceeding 
in  this  way,  we  may  reach  any  integer  n  we  may  mention, 
however  large  it  may  be.  Hence  (1)  is  true  [for  any  such 
value  of  n. 


XXII,  §  162]  BINOMIAL  THEOREM  257 

This  method  of  reasoning  illustrates  what  is  termed 
mathematical  induction.  Another  example  of  the  process 
will  now  be  given,  the  steps  being  arranged,  however,  in  a 
more  condensed  form. 

EXAMPLE.     Prove  by  mathematical  induction  that 

(1)  1+3+5+7'H  -----  \-(2n—  l)=n2.    (n  =  any  positive  integer) 

SOLUTION.  When  n  =  1,  the  formula  gives  1  =  I2  ;  when  n  =  2,  it 
gives  l+3=22;  when  n=3,  it  gives  l+3+5=32,  all  of  which 
arithmetical  relations  are  seen  to  be  correct. 

Let  r  represent  any  value  of  n  for  which  the  formula  has  been 
proved.  Then 

(2)  1+3+5+7+..  .  +(2  r-l)  =  r2. 

Adding  (2  r+1)  to  each  member  gives 

(3)  1+3+5+7+.-  -+(2  r+1)  =r2  +  (2  r  +  l)=r2+2  rfl  =  (r  +  l)2. 
But  (3)  is  the  same  as  (2)  except  that  r  has  been  replaced  through- 

out by  r  +  1.     Hence,  if  (1)  is  true  for  any  value  of  n,  such  as  r,  it 
is  necessarily  true  also  for  that  value  of  n  increased  by  1. 

Now,  we  know  (1)  to  be  true  when  n=3.  (See  above.)  Hence  it 
must  be  true  when  n  =  4.  Being  true  when  n  =  4,  it  must  be  true 
when  w=5,  etc.,  and  in  this  way  we  now  know  that  (1)  is  true  for 
any  value  (positive  integral)  of  n  whatever. 

EXERCISES 

Prove  the  correctness  of  each  of  the  following  formulas 
by  mathematical  induction,  n  always  being  understood  to 
be  any  positive  integer. 


[HINT.  First  try  out  f  orn  =  1,  n  =  2,  and  n  =  3.  Let  r  represent 
a  number  for  which  the  formula  holds.  Add  2(r  +  l)  to  both  mem- 
bers of  the  resulting  equation  and  compare  results.] 

2.   3+6+9+12+  •••+3  rc=  —  (n+1). 


3. 

4.   22+42+62+---+(2n)2  =£n(n+l)(2ri 
5. 

s 


258  SECOND   COURSE    IN   ALGEBRA     [XXII,  §  162 

6    _1 ,_1 i__l I [_       1       =    n 

1-2     2-3     3-4  n(w+l)     n+l' 

7.  2+22+23+24H r-2»  =  2(2»-l). 

8.  Prove  that  if  n  is  any  positive  integer,  an—bn  is  divisi- 
ble by  a  —  b. 

[HINT.  Since  or+1-6r+1  =a(ar -br)+br(a-b),  it  follows  that 
ar+1—  br+l  will  be  divisible  by  a  — 6  whenever  ar  —  br  is  divisible  by 
0-6.] 

9.  Prove  that  a2n  —  b2n  is  divisible  by  a +b. 

163.   The  Binomial  Theorem.     If  we  raise  the  binomial 
(a -\-x)  to  the  second  power,  that  is  find  (a+x)2,  the  result 
is  a2-}- 2  ax+x2  (§10).     Similarly,  by  repeated  multiplica- 
tion of  (a+x)  into  itself,  we  can  find  the  expanded  forms  for 
(a+z)3,  (a+x)4,  (a+x)5,  etc.     The  results  which  we  find  in 
this  way  have  been  placed  for  reference  in  a  table  below : 
>  =  a2+2ax+x2. 
'  =  a3+3  a2z+3  az2+z3. 
(ct-hx)4  =  tt4-|-4  a3x-\^6  a2x2-f-4  axs-}-x*. 
(a-f-x)5  =  tt5-|-6  a4x-j-10  a3x2-\-10  a2xs-\-Q  ax*-\-x5,  etc. 

Upon  comparing  these,  we  see  that  the  expansion  of  (a+x)n, 
where  n  is  any  positive  integer,  has  the  following  properties : 

1.  The  exponent  of  a  in  the  first  term  is  n,  and  it  decreases 
by  1  in  each  succeeding  term. 

The  last  term,  or  xn,  maybe  regarded  as  a°xn.     (See  §  122.) 

2.  The  first  term  does  not  contain  x.     The  exponent  of  x  in 
the  second  term  is  1  and  it  increases  by  1  in  each  succeeding 
term  until  it  becomes  n  in  the  last  term. 

3.  The  coefficient  of  the  first  term  is  1 ;     that  of  the  second 
term  is  n. 

4.  //  the  coefficient  of  any  term  be  multiplied  by  the  exponent 
of  a  in  that  term,  and  the  product  be  divided  by  the  number  of 
the  term,  the  quotient  is  the  coefficient  of  the  next  term. 


XXII,  §  163]  BINOMIAL  THEOREM  259 

For  example,  the  term  6  o2x2,  which  is  the  third  term  in  the  ex- 
pansion of  (o+z)4  (see  p.  258)  has  a  coefficient,  namely  6,  which 
may  be  derived  by  multiplying  the  coefficient  of  the  preceding  term 
(which  is  4)  by  the  exponent  of  a  in  that  term  (which  is  3)  and 
dividing  the  product  thus  obtained  by  the  number  of  that  term 
(which  is  2). 

5.    The  total  number  of  terms  in  the  expansion  is  n-\-  1. 

The  results  just  observed  regarding  the  expansion  of 
(a-\-x)n,  where  n  is  any  positive  integer,  may  be  summarized 
and  condensed  int.o  a  single  formula  as  follows  : 

(a+x)n  = 


1  •  2 


1-2-3 

the  dots  indicating  that  the  terms  are  to  be  supplied  in  the 
manner  indicated  up  to  the  last  one,  or  (w+l)st. 

This  formula  is  called  the  binomial  theorem.  By  means 
of  it,  one  may  write  down  at  once  the  expansion  of  any 
binomial  raised  to  any  positive  integral  power.  That  the 
formula  is  true  in  all  cases,  when  n  is  a  positive  integer,  will 
be  proved  in  detail  in  §  165.  We  assume  its  truth  here  for 
those  small  values  of  n  for  which  its  correctness  is  easily 
tested. 

NOTE.  The  formula  is  generally  attributed  to  Sir  Isaac  Newton 
(1642-1727)  ;  see  the  picture  facing  p.  193. 

EXAMPLE  1.     Expand  (a+z)6. 
SOLUTION.     Here  n  =  6,  so  the  formula  gives 

''' 


.6-5-4-3-2       5  .6-5-4-3.2-  1  x6 
^l-  2-  3-4-  5        ^1.2.3-4.5.6 

Simplifying  the  various  coefficients  by  performing  the  possible 
cancelations  in  each,  we  obtain 

5-Fz6.     Ans. 


260  SECOND    COURSE    IN   ALGEBRA     [XXII,  §  163 

NOTE.  It  may  be  observed  that  the  coefficients  of  the  first  and 
last  terms  turn  out  to  be  the  same  ;  likewise  the  coefficients  of  the 
second  and  next  to  the  last  terms  are  the  same,  and  so  on  symmetri- 
cally as  we  read  the  expansion  from  its  two  ends.  This  feature  is 
true  of  the  expansion  of  (a  +#)  to  any  power.  (Note  the  expansions 
of  (o+x)2,  (a-f-z)3,  (a+rc)4,  etc.,  as  given  at  the  beginning  of  §  163.) 

EXAMPLE  2.     Expand  (2  —  m)5. 

SOLUTION.  Here  a  =2,  x=—m,  and  n=5.  The  formula  thus 
gives 

S'  *'  j*  •  22(-m)3 

1  •  ^  •  o 


Simplifying  the  coefficients  (as  in  Example  1)  this  becomes 
(2-m)5=25+5-  24(-m)+10-  23(-m)2  +  10-  22(-ra)3 


Making  further  simplifications,  we  obtain 

(2-m)5  =  32-80m+80ra2-40ra3  +  10ra4-m5.     Ans. 


NOTE.  The  result  for  (2— x)5  is  the  same  as  that  for  (2+x)6 
except  that  the  signs  of  the  terms  are  alternately  positive  and 
negative  instead  of  all  positive.  A  similar  remark  applies  to  the 
expansion  of  every  binomial  of  the  form  (a—  x)n  as  compared  to 
that  of  (a+x)n. 

EXERCISES 

Expand  each  of  the  following  powers. 

1.  (x+y)*.  9.    (a2-z2)4.  1?     (I  ,  IV. 

2.  (a+6)4.  10.    (2a+l)4.  (x    yj 

3.  (x  —  vY.  11.    (x— 3  vY. 

40       /  u,       a/ 

4.  (a-6)4.  12.  (1+x2)6. 

5.  (2+r)5.  13.  (l-o:)8. 

6.  (a+x)7.  14.  (x  — ^-)5. 

7.  (fif-3)6.  15.  (3  a2- 1)4.  2a 


8.    (a2+x)6.  16.    (a+x) 


'» 


XXII,  §  1653  BINOMIAL  THEOREM  261 

*164.  The  General  Term  of  (a+x)n.  The  third  term  in  the 
expansion  of  (a+x)n,  as  given  by  the  formula  in  §  163,  is 

(n-l 

Observe  that  the  exponent  of  x  is  1  less  than  the  number  of  the 
term  ;  the  exponent  of  a  is  n  minus  the  exponent  of  x  ;  the  last 
factor  of  the  denominator  equals  the  exponent  of  x  ;  in  the  numerator 
there  are  as  many  factors  as  in  the  denominator. 

Precisely  the  same  statements  can  be  made  as  regards  the  fourth 
term,  or 


1  •  2  •  3 

In  the  same  way,  it  appears  that  the  above  statements  can  be 
made  of  any  term,  such  as  the  rth,  so  that  the  formula  for  the  rth 
term  is 

rth 


L  '  2i  '   O  •  •  •  (r  —  1) 

EXAMPLE.     Find  the  7th  term  of  (2  6  -c)10. 

SOLUTION.     Here  a  =  2  b,  x  =  (—c),  n  =  10,  and  r  =  7.     Therefore 
(using  the  formula),  the  desired  7th  term  is 

*!•  2-  3-  L  56   65  '  (26)4(-c)6  =  210(2&)4(-c)6  =  3360^c«.     Ans. 

EXERCISES 
Find  each  of  the  following  indicated  terms. 

1.  5th  term  of  («+*)-.  ?.    6th  term  of  tx+Vfl. 

2.  6th  term  of  (x-y}*. 

3.  7th  term  of  (2+  z)9.  8.    9th  term  of  ^- 

4.  10th  term  of  (w-n)14.  /2       2x12 
6.  6th  term  of  (*-&.)»  9'   5th  term  of  (f  -f  J    '_ 
6.  20th  term  of  (1  +x)24.  10.    4th  term  of  (2  \/2  -  v/3)  6. 
165.   Proof  of  the  Binomial  Theorem.     The  way  in  which 

the  binomial  formula  was  established  in  §  163  is,  strictly 
speaking,  open  to  objection  because  we  there  made  sure  of 
its  correctness  only  for  certain  special  values  of  n,  such  as 
ft  =  2,  n  =  3,  n  =  4,  and  n==5.  Though  the  formula  holds 


262  SECOND   COURSE    IN   ALGEBRA     [XXII,  §  165 

true,  as  we  saw,  in  these  cases,  it  does  not  follow  necessarily 
that  it  is  true  in  every  case,  that  is  for  eveiy  positive  inte- 
gral value  of  n.  We  can  now  establish  this  fact,  however, 
by  the  process  of  mathematical  induction,  when  n  is  a  positive 
integer. 

Let  ra  represent  any  special  value  of  n  for  which  the  formula 
has  been  established  (as,  for  example,  2,  3,  4,  or  5).  Then  we 
have 


(1)  m(m-l)  -  (m-r+2) 

1-2-3-  (r-1) 

Let  us  now  multiply  both  members  of  this  equation  by 
a+z.  On  the  left  we  obtain  (a+x)m+l.  On  the  right  we 
shall  have  the  sum  of  the  two  results  obtained  by  multiply- 
ing the  right  side  of  (1)  first  by  a  and  then  by  x,  that  is 
we  shall  have  the  sum  of  the  two  following  expressions  : 


m(m-l)  ...  (m-r+2) 

1-2.3-.  (r-1) 
and 


Adding  these,  and  making  the  natural  simplifications  in 
the  resulting  coefficients  of  amx,  am~lx2,  etc.,  and  equating 
the  final  result  to  its  equal  on  the  left  (namely  (a+x)m+1,  as 
noted  above)  gives 


(2) 


1  •  &  •  o  •••  (T  —  i) 


XXII,  §  167]  BINOMIAL  THEOREM  263 

But  (2)  is  precisely  (1)  except  for  the  substitution  of  ra+1 
for  m  throughout.  Hence,  if  the  binomial  formula  holds 
for  any  special  value  of  n,  as  m,  it  necessarily  holds  for  the 
next  larger  value,  namely  m-j-1.  But  we  have  already  ob- 
served that  it  holds  when  n  =  5.  It  must,  therefore,  hold  when 
n  =  5+l,  or  6.  But  if  it  holds  when  n  =  Q,  it  must  likewise 
hold  when  n  =  6+l,  or  7.  Thus  we  may  proceed  until  we 
arrive  at  any  chosen  value  of  n  whatever.  That  is,  the  for- 
mula must  be  true  for  any  positive  integral  value  of  n. 

*166.  The  Binomial  Formula  for  Fractional  and  Negative  Ex- 
ponents. In  case  the  exponent  n  is  not  a  positive  integer  but  is 
fractional  or  negative,  we  may  still  write  the  expansion  of  (a-\-x)n 
by  the  formula  of  §  163,  but  it  will  now  contain  indefinitely  many 
terms  instead  of  coming  to  an  end  at  some  definite  point,  that  is 
we  meet  with  an  infinite  series.  (Compare  §  92.) 

For  example,  the  formula  gives 


1-2  1  •  2  •  3 


=  a1/2  +4-  a-1/2.?  +n      a~3/2x2  -f~~      a-5/2x3  +• 
1-2  1-2-3 


=  a^+±a-^x-±a-*< 

Here  we  have  written  only  the  first  four  terms  of  the  expansion, 
but  we  could  obtain  the  5th  term  in  the  same  way  and  as  many 
others  in  their  order  as  might  be  desired. 

*167.  Application.  If  in  (a+x}n  the  value  of  x  is  small  in  com- 
parison to  that  of  a  (more  exactly,  if  the  numerical  value  of  x/a  is 
less  than  1)  then  the  first  few  terms  of  the  expansion  furnish  a 
close  approximation  to  the  value  of  (a+z)n.  This  fact  is  often 
used  to  find  approximate  values  for  the  roots  of  numbers  in  the 
manner  illustrated  below. 

EXAMPLE.     Find  the  approximate  value  of  VlO. 

SOLUTION.  Write  VlO  =  V9  +  l  =  V(32  +  l)  and  expand  this 
last  form  by  the  binomial  formula.  Thus  (using  the  final  result 


264  SECOND   COURSE    IN   ALGEBRA      [XXII,  §  167 

in  the  worked  example  of  §  166),  we  have 

1/2-  1--  J-(32)-3/2-  I2 


o  ,_L_      1.1 


2-3    8-33     16-35 
=  3+.166666-.004629+.000257  =3.162288  (approximately). 

Observe  that  the  value  of  VlO  as  given  in  the  tables  is  3.16228, 
thus  agreeing  with  that  just  found  so  far  as  the  first  five  places  of 
decimals  are  concerned. 

Whenever  extracting  roots  by  this  process  we  use  the  following 
general  rule. 

Separate  the  given  number  into  two  parts,  the  first  of  which  is  the 
nearest  perfect  power  of  the  same  degree  as  the  required  root,  and  ex- 
pand the  result  by  the  binomial  theorem. 

*EXERCISES 

Write  the  first  four  terms  in  the  expansion  of  each  of  the  following 
expressions. 

1.  (o+aO»/».  6.    (2  o+6)3/4. 

2.  (a+z)-2.  6.    (a3-z2)-3/4. 

3.  (l+x)V3.  7. 

4.  (2-  a;)-1/4.  8. 

9.   Find  the  6th  term  in  the  expansion  of 

[HINT.     Use  the  formula  in  §  164,  with  n  =-J  and  r  =  6.] 

Find  the 

10.  5th  term  of  (a+rr)1/2.  13.  9th  term  of  (a  - 

11.  7th  term  of  (a+x)~2/3.  14.  10th  term  of 

12.  8th  term  of  (1+z)1/3.  15.  6th  term  of  \/2a+6. 

Find  the  approximate  values  of  the  following  to  six  decimal 
places  and  compare  your  results  for  the  first  three  examples  with 
those  given  in  the  tables. 

16.    VT7.  17.    V27.  18.    \/9 

19.    ^11.  20. 

[HINT.    Write  14  =  16-2=24-2.] 


CHAPTER  XXIII 


THE   SOLUTION  OF  EQUATIONS  BY    DETERMINANTS 
168.   Definitions.    The  symbol 


a    b 
c    d 


is  called  a  determinant  of  the  second  order,  and  is  defined 
as  follows  : 

a    b 


Thus 


8    3 
2    4 


7 
2 

-10 

Q 


=  ad—bc. 


8  -4-2  -3  =  32-6  =  26. 


=  7  -4-(-2)  -3  =  28+6  =  34. 


6]  =  4[-50+18] 
=  4(-32)  =  -128. 


The  numbers  a,  b,  c,  and  d  are  called  the  elements  of  the 
determinant. 

The  elements  a  and  d  (which  lie  along  the  diagonal  through 
the  upper  left-hand  corner  of  the  determinant)  form  the 
principal  diagonal  The  letters  b  and  c  (which  lie  along  the 
diagonal  through  the  upper  right-hand  corner)  form  the 

minor  diagonal. 

265 


266 


SECOND   COURSE    IN   ALGEBRA     [XXIII,  §  168 


From  these  definitions,  we  have  the  following  rule. 

To  evaluate  any  determinant  of  the  second  order,  subtract 
the  product  of  the  elements  in  the  minor  diagonal  from  the 
product  of  the  elements  in  the  principal  diagonal. 

EXERCISES 

Evaluate  each  of  the  following  determinants. 


1. 


3. 


7        1 
4     -8 

8    7 
-2    3 

-2        6 

7     -3 


6.    5 


7.    f 


8. 


2a 
3a 


7a 


36 
56 


0 
66 


x+y    3 
x  —  y    4 


169.  Solution  of  Two  Linear  Equations.  Let  us  consider 
a  system  of  two  linear  equations  between  two  unknown 
letters,  x  and  y.  Any  such  system  is  of  the  form 

(1) 

(2) 


where  ai,  61,  Ci,  etc.,  represent  known  numbers  (coefficients). 
This  system  may  be  solved  for  x  and  y  by  elimination,  as 
in  Chapter  VII.  Thus,  multiplying  (1)  by  62  and  (2)  by  61 
and  then  subtracting  the  resulting  equations  from  each  other, 
the  letter  y  is  eliminated  and  we  reach  the  equation 

(«i62  —  0^61)0;  =  62Ci  —  bic%. 
Therefore 

(3)  x 


Likewise,  we  may  eliminate  x  by  multiplying  (1)  by  02  and 


XXIII,  §  169] 


DETERMINANTS 


267 


(2)  by  «i  and  subtracting  the  resulting  equations  from  each 
other.     This  gives 

(ai&2  —  0261)  y  =  aid  — 
Therefore 


(4) 


y 


It  is  now  clear,  by  §  168,  that  the  numerators  and  denomi- 
nators in  (3)  and  (4)  are  all  determinants  of  the  second  order ; 
and  by  the  definition  of  §  168,  (3)  and  (4)  may  be  written 
respectively  in  the  forms 

61 


(5) 


x  = 


y= 


These  forms  for  the  solution  of  (1)  and  (2)  are  easily  re- 
membered. In  particular,  observe  that : 

1.  The  determinant  for  the  denominator  is  the  same  for 
both  x  and  y. 

2.  The  determinant  for  the  numerator  of  the  z-value  is  the 
same  as  that  for  the  denominator  except  that  the  numbers 
Ci  and  c%  replace  the  ai  and  a^  which  occur  in  the  first  column 
of  the  denominator  determinant. 

3.  The  determinant  for  the  numerator  of  the  ^/-value  is 
the  same  as  that  for  the  denominator  except  that  the  num- 
bers Ci  and  02  replace  the  bi  and  62  which  occur  in  the  second 
column  of  the  denominator  determinant. 

The  usefulness  of  the  forms  (5)  lies  in  the  fact  that  they 
express  the  solution  of  a  system  of  two  linear  equations  in 
condensed  form,  enabling  us  to  write  down  the  desired  values 
of  x  and  y  immediately,  without  the  usual  process  of  elimina- 
tion. This  will  now  be  illustrated. 


268  SECOND   COURSE    IN   ALGEBRA     [XXIII,  §  169 

EXAMPLE.     Solve  by  determinants  the  system 


(6) 

(7)  x-7y=-S. 

SOLUTION.     Using  the  forms  (5),  we  have  at  once 


x  = 


18 

3 

-8 

-7 

_18.  (_7)-(-8)- 

3     -126+24     -102     R 

2 

3 

2(-7)-l  •  3 

-14-3         -17 

1 

-7 

2 

18 

1 

—8 

2-  (-8)-!-  18 

-16-18     -34     o 

2 

3 

2(—  7)—  1  •  3 

-14-3      -17 

1 

-7 

The  solution  desired  is  therefore  (x  =  6,  y=2).     Ans. 

CHECK.     Substituting  6 'for  #  and  2  for  y  in  (6)  and  (7)  gives 
12  +6  =  18  and  6  - 14  =  -8,  which  are  true  results. 

EXERCISES 

Solve  each  of  the  following  pairs  of  equations  by  determi- 
nants, checking  your  answers  for  each  of  the  first  three. 


1. 
2. 
3. 
4. 


Sx+3y=-7. 

3x-7  y=-&, 
=  14. 


7. 


8. 


I  2x-y=lS. 

I  ax-\-by  =  r, 
\bx  —  ay  =  s. 


j  .2a4-.56  =  30, 
5*    \. 4  a-. 8  6=  -16. 


XXIII,  §  170] 


DETERMINANTS 


269 


170.    Determinants  of  the  Third  Order.     The  symbol 


(1) 


is  called  a  determinant  of  the  third  order. 

Its  value  is  defined  as  follows  : 

This  expression,  as  we  shall  see  presently,  is  important  in  the 
study  of  equations. 

The  expression  (2)  is  called  the  expanded  form  of  the  determinant 
(1).  It  is  important  to  observe  that  this  expanded  form  may  be 
written  down  at  once  as  follows. 

Write  the  determinant  with  the  first  two  columns  repeated  at 
the  right  and  first  note  the  three  diagonals  which  then  run  down 
from  left  to  right  (marked  +).  The 
product  of  the  elements  in  the  first  of 
these  diagonals  is  ai  62  c3,  and  this  is  seen  to 
be  the  first  term  of  the  expanded  form  (2). 
Similarly,  the  product  of  the  elements  in 
the  second  of  these  diagonals  is  6iC2«3, 
which  forms  the  second  term  of  (2) ;  and 
likewise  the  third  diagonal  furnishes  at 
once  the  third  term  of  (2). 

Next  consider  the  three  diagonals  which  run  up  from  left  to  right 
(marked  with  dotted  lines) .  The  product  of  the  elements  in  the  first 
of  these  is  o3  62  Ci,  and  this  is  the  fourth  term  of  (2),  provided  it  be 
taken  negatively,  that  is  preceded  by  the  sign  — .  Similarly,  the 
other  two  dotted  diagonals  of  (3)  furnish  the  last  two  terms  of  (2), 
provided  they  be  taken  negatively. 

NOTE.  Every  determinant  of  the  third  order  when  expanded 
contains  a  total  of  six  terms. 

EXAMPLE.     Expand  and  find  the  value  of  the  determinant 


379 
214 
632 


270 


SECOND   COURSE    IN   ALGEBRA    [XXIII,  §  170 


SOLUTION.     Repeating  the  first  and  second  columns  at  the  right, 
we  have 


379 
214 
632 


3  7 
2  1 
6  3 


The  diagonals  running  down  from  left  to  right  give  the  three 
products 

3-1-2,     7-4-6,     9-2-3, 

which  form  the  first  three  terms  of  the  expansion. 

The  diagonals  running  up  from  left  to  right  give  the  products 

6-1-9,     3-4-3,     2-2-7, 

which,  when  taken  negatively,  form  the  three  remaining  terms  of  the 
determinant. 

The  complete  expanded  form  of  (3)  is,  therefore, 

3-  1- 2+7-4- 6+9-2- 3-6- -1-9-3-4. 3-2- 2- 7, 
which  reduces  to 

6+168+54-54-36-28  =  110.     Ans. 


*  EXERCISES 
Expand  and  find  the  value  of  the  following  determinants. 


3. 


4. 


13  7 
24  6 
35-4 


-7 
3 

2         2 
-4        6 

8 

-5     -3 

8 

2 

3 

6 

0 

5 

. 

3 

0 

7 

2a    3        66 

3  a     2     -56 

a.  0     -26 


5. 


6. 


7. 


8. 


x  7  1 
23-4 
4  2  1 

a     6     2 

-453 

210 

a     6     c 
d     e     f 

x     y     z 

1  0  0 
0  x-y  0 
0  0  x+y 


XXIII,  §  171] 


DETERMINANTS 


271 


*171.  Solution  of  Three  Linear  Equations.  Let  us  consider  a  sys- 
tem of  three  linear  equations  between  three  unknown  letters,  such 
as  x,  y,  and  z.  Any  such  system  is  of  the  form 


(1) 


where  01,  61,  ci,  di,  a2,  b2,  etc.,  represent  known  numbers  (coefficients). 
This  system  may  be  solved  for  x,  y,  and  z  by  elimination,  as  in  §  35, 
but  the  process  is  long.     We  shall  here  state  merely  the  results, 
which  are  as  follows  (compare  with  (3)  and  (4)  of  §  169)  : 

_dibzC3-\-d2b3c\-}-dzbiCz  —  dsbzCi 


_ 


gib-ids  -\-dzb3di  -\-d3bidz—d3bzdi  —  dib3dz—a3bidzw 


(2) 


It  is  clear  by  §  170  that  in  these  values  for  x,  y,  and  z,  each  nu- 
merator and  denominator  is  the  expanded  form  of  a  determinant  of 
the  third  order.  In  fact,  it  appears  from  the  definition  in  §  170, 
that  we  may  now  express  these  values  of  x,  y,  and  z  in  the  following 
condensed  (determinant)  forms : 

i     61     ( 


(3)    x  = 


61 


b3 


61 


d3 


The  importance  of  these  expressions  for  x,  y,  and  z  lies  in  the  fact 
that  they  give  at  once  the  solution  of  any  system  such  as  (1)  in 
very  compact  and  easily  remembered  forms.  The  following  features 
should  be  especially  noted  : 

1.  The  denominator  determinant  is  the  same  in  all  three  cases. 
(Compare  statement  1  of  §  169.) 

2.  The  determinant  for  the  numerator  of  the  z-value  is  the  same 
as  that  for  the  denominator  determinant  except  that  the  numbers 
di,  dz,  dz  replace  the  01,  a2,  03  which  occur  in  the  first  column  of  the 
denominator  determinant. 


272 


SECOND   COURSE    IN   ALGEBRA     [XXIII,  §  171 


3.  Similarly,  the  numerator  of  the  y-v&lue  is  formed  from  that 
of  the  denominator  determinant  by  replacing  the  second  column  by 
the  elements  di,  d2,  d3 ;  while  the  numerator  of  the  2-value  is  formed 
from  that  of  the  denominator  determinant  by  replacing  the  third 
column  by  the  elements  dif  d2,  d3.  (Compare  statements  2  and  3 
of  §  169.) 

The  readiness  with  which  (3)  may  be  used  in  practice  to  solve  a 
system  of  three  linear  equations  is  illustrated  by  the  following 

EXAMPLE.     Solve  the  system 

2x-y+3  0  =  35, 
z+37/-15=-2z, 

SOLUTION.  Arranging  the  equations  as  in  (1)  of  §  171,  the  given 
system  is 

2  x-y  +3  z  =35, 


Therefore,  using  (3)  of  §  171,  we  have  at  once 


0+180-2-9-280-0_-lll_ 
:  0  +  12-6-27-16-0  ~~^ 


35 

-1 

3 

15 

3 

2 

1 

4 

0 

2 

-1 

3 

1 

3 

2 

3 

4 

0 

2 

35 

3 

1 

15 

2 

3 

1 

0 

2 

-1 

3 

1 

3 

2 

3 

4 

0 

2 

-1 

35 

1 

3 

15 

3 

4 

1 

2 

-1 

3 

1 

3 

2 

3 

4 

0 

_o      ,R 
~3' 


_0+3+210-135-4-Q^   74 
-37  -37 


=  -2, 


_6  +  140-45-315-120+l^-333 
-37  "  -37 


=  9. 


XXIII,  §  172]  DETERMINANTS  273 

The  desired  solution  is,  therefore,  (3=3,  0=  —2,  2  =  9).     Ans. 
CHECK.     With  x  =  3,  y  =  —2,  z  =  9,  it  is  readily  seen  that  the  three 
given  equations  are  satisfied. 


EXERCISES 


Solve  each  of  the  following  systems  by  determinants. 

3-20+2=5,  f  3+20=0, 

1.     {     33+60-42=3,  6.  3+2= -3, 

83-100+32=34.  [40+3-22  =  4. 


f  23-3  0-42  =  25, 
3.  3 +0-2  =-4, 


4. 


-9»  f  3+0  =  3  a, 

33+20-2  =  2,  7.     <|z+2=56, 

23-0+2=^. 


*  172.  Determinants  of  Higher  Order.  Determinants  of  the 
fourth  order  exist  and  are  studied  in  higher  algebra,  as  are  deter- 
minants of  the  fifth  order,  sixth  order,  etc.  Moreover,  determi- 
nants of  the  fourth  order  bear  a  similar  relation  to  the  solving  of 
four  linear  equations  between  four  unknown  letters,  as  determinants 
of  the  third  order  bear  to  the  solving  of  three  linear  equations  be- 
tween three  unknown  letters ;  and  a  similar  remark  may  be  made 
regarding  determinants  of  the  fifth  order,  sixth  order,  etc.  In  all 
cases,  the  solutions  of  such  systems  of  equations  can  be  expressed 
very  simply  by  means  of  determinants. 


APPENDIX 

TABLE  OF  POWERS  AND  ROOTS 
EXPLANATION 

1.  Square  Roots.  The  way  to  find  square  roots  from  the 
Table  is  best  understood  from  an  example.  Thus,  suppose 
we  wish  to  find  Vl.48.  To  do  this  we  first  locate  1.48  in 
the  column  headed  by  the  letter  n.  We  find  it  near  the 
bottom  of  this  column  (next  to  the  last  number).  Now 
we  go  across  on  that  level  until  we  get  into  the  column 
headed  by  Vn.  We  find  at  that  place  the  number  1.21655. 
This  is  .our  answer.  That  is,  Vl.48  =1.21655  (approxi- 
mately) . 

If  we  had  wanted  V14.8  instead  of  Vl.48  the  work  would 
have  been  the  same  except  that  we  would  have  gone  over 
into  the  column  headed  VlO  n  (because  14.8=10X1.48). 
The  number  thus  located  is  seen  to  be  3.84708,  which  is, 
therefore,  the  desired  value  of  V14.8. 

Again,  if  we  had  wished  to  find  Vl48  the  work  would  take 
us  back  again  to  the  column  headed  Vn,  but  now  instead 
of  the  answer  being  1.21655  it  would  be  12.1655.  In  other 
words,  the  order  of  the  digits  in  Vl48  is  the  same  as  for 
Vl.48,  but  the  decimal  point  in  the  answer  is  one  place 
farther  to  the  right. 

Similarly,  if  we  desired  V1480  the  work  would  be  the  same 
as  before  except  that  we  must  now  use  the  column  headed 
VlO  n  and  move  the  decimal  point  there  occurring  one  place 
farther  to  the  right.  This  is  seen  to  give  38.4708. 

Thus  we  see  how  to  get  the  square  root  of  1.48  or  any 
power  of  10  times  that  number. 

275 


276  APPENDIX 

In  the  same  way,  if  we  wish  to  find  V.148,  or  V.0148,  or 
V.00148,  or  the  square  root  of  any  number  obtained  by 
dividing  1.48  by  any  power  of  10  we  can  get  the  answers 
from  the  column  headed  Vn  or  VlOn  by  merely  placing 
the  decimal  point  properly.  Thus,  we  find  that  V.148  = 
.384708,  V^148  =  .121655,  VML48  =  .0384708,  etc. 

What  we  have  seen  in  regard  to  the  square  root  of  1.48 
or  of  that  number  multiplied  or  divided  by  any  power  of  10 
holds  true  in  a  similar  way  for  any  number  that  occurs  in 
the  column  headed  n,  so  that  the  tables  thus  give  us  the 
square  roots  of  a  great  many  numbers. 

2.  Cube  Roots.  Cube  roots  are  located  in  the  tables 
in  much  the  same  way  as  that  just  described  for  square 
roots,  but  we  have  here  three  columns  to  select  from  instead 
of  two,  namely  the  columns  headed  Vn,  VlO  n,  VlOO  n. 

Illustration. 

•v/1.48  occurs  in  the  column  headed  %/n  and  is  seen  to  be  1.13960. 
-v/14.8  occurs  in  the  column  headed  -^10  n  and  is  seen  to  be  2.4552. 
\/148  occurs  in  the  column  headed    v'lOO  n  and  is  seen  to  be 
5.28957. 

To  get  vO48  we  observe  that  .148  =  ^M^  =  ^jjj:  =  ^  ^148. 

*    .LvJ  JLLHJU        -L\J 

Thus,  we  look  up  \/148  and  divide  it  by  10.    The  result  is  instantly 
seen  to  be  .528957.    Similarly,   to  get   \/.0148  we  observe  that 

^0148  =  -vS|  =  A/J^  =  i  3/Ul8.    Thus,  we  look  up  \/14^Fand 
10U  •        1UUU      10 

divide  it  by  10,  giving  the  result  .24552. 

To  get  ^.00148  we  observe  that  V/.00148  =  A/T?H  =  A  V^8'  so 

-LvJv/vJ       JLU 

that  we  must  divide  Vl.48  by  10.    This  gives  .11396. 

Similarly  the  cube  root  of  any  number  occurring  in  the 
column  headed  n  may  be  found,  as  well  as  the  cube  root  of 
any  number  obtained  by  multiplying  or  dividing  such  a 
number  by  any  power  of  10. 


TABLE   OF  POWERS  AND   ROOTS  277 

3.  Squares  and  Cubes.  To  find  the  square  of  1.48  we 
naturally  look  at  the  proper  level  in  the  column  headed  n2. 
Here  we  find  2.1904,  which  is  the  answer.  If  we  wished  the 
square  of  14.8  the  result  would  be  the  same  except  that 
the  decimal  point  must  be  moved  two  places  to  the  right, 
giving  219.04  as  the  answer.  Similarly  the  value  of  (148)2 
is  21904.0  etc. 

On  the  other  hand,  the  value  of  (.148)2  is  found  by  moving 
the  decimal  place  two  places  to  the  left,  thus  giving  .021904. 
Similarly,  (.0148)2  =  .00021904,  etc. 

To  find  (1.48)3  we  look  at  the  proper  level  in  the  column 
headed  n3  where  we  find  3.24179.  The  value  of  (14.8)3  is 
the  same  except  that  we  must  move  the  decimal  point  three 
places  to  the  right,  giving  3241.79.  Similarly,  in  finding 
(.148)3  we  must  move  the  decimal  place  three  places  to  the 
left,  giving  .00324179. 

Further  illustrations  of  the  way  to  use  the  tables  will  be 
found  in  §  43. 

EXERCISES 

Read  off  from  the  tables  the  values  of  each  of  the  following  ex- 
pressions. 

1.    \/4l  4.    v'GTO  7.    V93/7  10.    v/,00154 


2.  V8.9  5.    V^9  8.    V93.7  11.    V.000143 

3.  >/67  Q.    VjOlG  9.    V.00154         12.    v^.000143 


278 


Table  I  — Powers  and  Roots 


n 

n2 

V^ 

vio™ 

n* 

*£ 

VWn 

^ioow 

1.00 

1.0000 

1.00000 

3.16228 

1.00000 

1.00000 

2.15443 

4.64159 

1.01 
1.02 
1.03 

1.0201 
1.0404 
1.0609 

1.00499 
•  1.00995 
1.01489 

3.17805 
3.19374 
3.20936 

1.03030 
1.06121 
1.09273 

1.00332 
1.00662 
1.00990 

2.16159 
2.16870 
2.17577 

4.65701 
4.67233 
4.68755 

1.04 
1.05 
1.06 

1.0816 
1.1025 
1.1236 

1.01980 
1.02470 
1.02956 

3.22490 
3.24037 
3.25576 

1.12486 
1.15762 
1.19102 

1.01316 
1.01640 
1.01961 

2.18279 
2.18976 
2.19669 

4.70267 
4.71769 
4.73262 

1.07 

1.08 
1.09 

1.1449 
1.1664 
1.1881 

1.03441 
1.03923 
1.04403 

3.27109 
3.28634 
3.30151 

1.22504 
1.25971 
1.29503 

1.02281 
1.02599 
1.02914 

2.20358 
2.21042 
2.21722 

4.74746 
4.76220 
4.77686 

1.10 

1.2100 

1.04881 

3.31662 

1.33100 

1.03228 

2.22398 

4.79142 

1.11 
1.12 
1.13 

1.2321 
1.2544 
1.2769 

1.05357 
1.05830 
1.06301 

3.33167 
3.34664 
3.36155 

1.36763 
1.40493 
1.44290 

1.03540 
1.03850 
1.04158 

2.23070 
2.23738 
2.24402 

4.80590 
4.82028 
4.83459 

1.14 
.15 
.16 

1.2996 
1.3225 
1.3456 

1.06771 
1.07238 
1.07703 

3.37639 
3.39116 
3.40588 

1.48154 
1.52088 
1.56090 

1.04464 
1.04769 
1.05072 

2.25062 
2.25718 
2.26370 

4.84881 
4.86294 
4.87700 

.17 

.18 
.19 

1.3689 
1.3924 
1.4161 

1.08167 
1.08628 
1.09087 

3.42053 
3.43511 
3.44964 

1.60161 
1.64303 
1.68516 

1.05373 

1.05672 
1.05970 

2.27019 
2.27664 
2.28305 

4.89097 
4.90487 
4.91868 

1.20 

1.4400 

1.09545 

3.46410 

1.72800 

1.06266 

2.28943 

4.93242 

1.21 
1.22 
1.23 

1.4641 
1.4884 
1.5129 

1.10000 
1.10454 
1.10905 

3.47851 
3.49285 
3.50714 

1.77156 
1.81585 
1.86087 

1.06560 
1.06853 
1.07144 

2.29577 
2.30208 
2.30835 

4.94609 
4.95968 
4.97319 

1.24 
1.25 
1.26 

1.5376 
1.5625 
1.5876 

1.11355 
1.11803 
1.12250 

3.52136 
3.53553 
3.54965 

1.90662 
1.95312 
2.00038 

1.07434 
1.07722 
1.08008 

2.31459 
2.32079 
2.32697 

4.98663 
5.00000 
5.01330 

1.27 
1.28 
1.29 

1.6129 
1.6384 
1.6641 

1.12694 
1.13137 
1.13578 

3.56371 
3.57771 
3.59166 

2.04838 
2.09715 
2.14669 

1.08293 
1.08577 
1.08859 

2.33311 
2.33921 
2.34529 

5.02653 
5.03968 
5.05277 

1.30 

1.6900 

1.14018 

3.60555 

2.19700 

1.09139 

2.35133 

5.06580 

1.31 
1.32 
1.33 

1.7161 
1.7424 
1.7689 

1.14455 
1.14891 
1.15326 

3.61939 
3.63318 
3.64692 

2.24809 
2.29997 
2.35264 

1.09418 
1.09696 
1.09972 

2.35735 
2.36333 
2.36928 

5.07875 
5.09164 
5.10447 

1.34 
1.35 
1.36 

1.7956 
1.8225 
1.8496 

1.15758 
1.16190 
1.16619 

3.66060 
3.67423 
3.68782 

2.40610 
2.46038 
2.51546 

1.10247 
1.10521 
1.10793 

2.37521 
2.38110 
2.38697 

5.11723 
5.12993 
5.14256 

1.37 
1.38 
1.39 

1.8769 
1.9044 
1.9321 

1.17047 
1.17473 
1.17898 

3.70135 

3.71484 
3.72827 

2.57135 
2.62807 
2.68562 

1.11064 
1.11334 
1.11602 

2.39280 
2.39861 
2.40439 

5.15514 
5.16765 
5.18010 

1.40 

1.9600 

1.18322 

3.74166 

2.74400 

1.11869 

2.41014 

5.19249 

1.41 
1.42 
1.43 

1.9881 
2.0164 
2.0449 

1.18743 
1.19164 
1.19583 

3.75500 
3.76829 
3.78153 

2.80322 
2.86329 
2.92421 

1.12135 
1.12399 
1.12662 

2.41587 
2.42156 
2.42724 

5.20483 
5.21710 
5.22932 

1.44 
1.45 
1.46 

2.0736 
2.1025 
2.1316 

1.20000 
1.20416 
1.20830 

3.79473 
3.80789 
3.82099 

2.98598 
3.04862 
3.11214 

1.12924 
1.13185 
1.13445 

2.43288 
2.43850 
2.44409 

6.24148 
5.26359 
6.26564 

1.47 
1.48 
1.49 

2.1609 
2.1904 
2.2201 

1.21244 
1.21655 
1.22066 

3.83406 
3.84708 
3.86005 

3.17652 
3.24179 

3.30795 

1.13703 
1.13960 
1.14216 

2.44966 
2.46620 

±•111071! 

6.27763 

5.28957 
6.30146 

Powers  and  Roots 


279 


n 

nz 

Vn 

VlOw 

nB 

Vn 

^10  n 

VWQn 

1.50 

2.2500 

1.22474 

3.87298 

3.37500 

1.14471 

2.46621 

5.31329 

1.51 
1.52 
1.53 

2.2801 
2.3104 
2.3409 

1.22882 
1.23288 
1.23693 

3.88587 
3.89872 
3.91152 

3.44295 
3.51181 
3.58158 

1.14725 
1.14978 
1.15230 

2.47168 
2.47712 

2.48255 

5.32507 
5.33680 
5.34848 

1.54 
1.55  • 
1.56 

2.3716 
2.4025 
2.4336 

1.24097 
1.24499 
1.24900 

3.92428 
3.93700 
3.94968 

3.65226 
3.72388 
3.79642 

1.15480 
1.15729 
1.15978 

2.48794 
2.49332 
2.49867 

5.36011 
5.37169 
5.38321 

1.57  ; 
1.58 
1.59 

2.4649 
2.4964 
2.5281 

1.25300 
1.25698 
1.26095 

3.96232 
3.97492 

3.98748 

3.86989 
3.94431 
4.01968 

1.16225 
1.16471 
1.16717 

2.50399 
2.50930 
2.51458 

5.39469 
5.40612 
5.41750 

1.60 

2.5600 

1.26491 

4.00000 

4.09600 

1.16961 

2.51984 

5.42884 

1.61 
1.62 
1.63 

2.5921 
2.6244 
2.6569 

1.26886 
1.27279 
1.27671 

4.01248 
4.02492 
4.03733 

4.17328 
4.25153 
4.33075 

1.17204 
1.17446 
1.17687 

2.52508 
2.53030 
2.53549 

5.44012 
5.45136 
5.46256 

1.64 
1.65 
1.66 

2.6896 
2.7225 
2.7556 

1.28062 
1.28452 
1.28841 

4.04969 
4.06202 
4.07431 

4.41094 
4.49212 
4.57430 

1.17927 
1.18167 
1.18405 

2.54067 
2.54582 
2.55095 

5.47370 
5.48481 
5.49586 

1.67 
1.68 
1.69 

2.7889 
2.8224 
2.8561 

1.29228 
1.29615 
1.30000 

4.08656 
4.09878 
4.11096 

4.65746 
4.74163 

4.82681 

1.18642 
1.18878 
1.19114 

2.55607 
2.56116 
2.56623 

5.50688 
5.51785 
5.52877 

1.70 

2.8900 

1.30384 

4.12311 

4.91300 

1.19348 

2.57128 

5.53966 

1.71 
1.72 
1.73 

2.9241 
2.9584 
2.9929 

1.30767 
1.31149 
1.31529 

4.13521 
4.14729 
4.15933 

5.00021 
5.08845 
5.17772 

1.19582 
1.19815 
1.20046 

2.57631 
2.58133 
2.58632 

5.55050 
5.56130 
5.57205 

1.74 
1.75 
1.76 

3.0276 
3.0625 
3.0976 

1.31909 
1.32288 
1.32665 

4.17133 
4.18330 
4.19524 

5.26802 
5.35938 
5.45178 

1.20277 
1.20507 
1.20736 

2.59129 
2.59625 
2.60118 

5.58277 
5.59344 
5.60408 

1.77 
1.78 
1.79 

3.1329 
3.1684 
3.2041 

1.33041 
1.33417 
1.33791 

4.20714 
4.21900 
4.23084 

5.54523 
5.63975 
5.73534 

1.20964 
1.21192 
1.21418 

2.60610 
2.61100 
2.61588 

5.61467 
5.62523 
5.63574 

1.80 

3.2400 

1.34164 

4.24264 

5.83200 

1.21644 

2.62074 

5.64622 

1.81 
1.82 
1.83 

3.2761 
3.3124 
3.3489 

1.34536 
1.34907 
1.35277 

4.25441 
4.26615 
4.27785 

5.92974 
6.02857 
6.12849 

1.21869 
1.22093 
1.22316 

2.62559 
2.63041 
2.63522 

5.65665 
5.66705 
5.67741 

1.84 
1.85 
1.86 

3.3856 
3.4225 
3.4596 

1.35647 
1.36015 
1.36382 

4.28952 
4.30116 
4.31277 

6.22950 
6.33162 
6.43486 

1.22539 

1.22760 
1.22981 

2.64001 
2.64479 
2.64954 

5.68773 
5.69802 
5.70827 

1.87 
1.88 
1.89 

3.4969 
3.5344 
3.5721 

1.36748 
1.37113 
1.37477 

4.32435 
4.33590 
4.34741 

6.53920 
6.64467 
6.75127 

1.23201 
1.23420 
1.23639 

2.65428 
2.65901 
2.66371 

5.71848 
5.72865 
5.73879 

1.90 

3.6100 

1.37840 

4.35890 

6.85900 

1.23856 

2.66840 

5.74890 

1.91 
1.92 
1.93 

3.6481 
3.6864 
3.7249 

1.38203 
1.38564 
1.38924 

4.37035 
4.38178 
4.39318 

6.96787 
7.07789 
7.18906 

1.24073 
1.24289 
1.24505 

2.67307 
2.67773 
2.68237 

5.75897 
5.76900 
5.77900 

1.94 
1.95 
1.96 

3.7636 
3.8025 
3.8416 

1.39284 
1.39642 
1.40000 

4.40454 
4.41588 
4.42719 

7.30138 
7.41488 
7.52954 

1.24719 
1.24933 
1.25146 

2.68700 
2.69161 
2.69620 

5.78896 
5.79889 
5.80879 

1.97 

1.98 
1.99 

3.8809 
3.9204 
3.9601 

1.40357 
1.40712 
1.41067 

4.43847 
4.44972 
4.46094 

7.64537 
7.76239 
7.88060 

1.25359 
1.25571 
1.25782 

2.70078 
2.70534 
2.70989 

5.81865 
5.82848 
5.83827 

280 


Powers  and  Roots 


n 

n2 

v^ 

vio™ 

n8 

tfi 

^Wn 

VWOn 

2.00 

4.0000 

1.41421 

4.47214 

8.00000 

1.25992 

2.71442 

5.84804 

2.01 
2.02 
2.03 

4.0401 
4.0804 
4.1209 

1.41774 
1.42127 

1.42478 

448330 
4.49444 
4.50555 

8.12060 
8.24241 
8.36543 

1.26202 
1.26411 
1.26619 

2.71893 
2.72344 
2.72792 

5.85777 
5.86746 
5.87713 

2.04 
2.05 
2.06 

4.1616 
4.2025 
4.2436 

1.42829 
1.43178 
1.43527 

4.51664 
4.52769 
4.53872 

8.48966 
8.61512 

8.74182 

1.26827 
1.27033 
1.27240 

2.73239 
2.73685 
2.74129 

5.88677 
6.89637 
5.90594 

2.07 
2.08 
2.09 

4.2849 
4.3264 
4.3681 

1.43875 
1.44222 
1.44568 

4.54973 
4.56070 
4.57165 

8.86974 
8.99891 
9.12933 

1.27445 
1.27650 
1.27854 

2.74572 
2.75014 
2.75454 

5.91548 
5.92499 
5.93447 

2.10 

4.4100 

1.44914 

4.58258 

9.26100 

1.28058 

2.75892 

5.94392 

2.11 
2.12 
2.13 

4.4521 

4.4944 
4.5369 

1.45258 
1.45602 
1.45945 

4.59347 
4.60435 
4.61519 

9.39393 
9.52813 
9.66360 

1.28261 
1.28463 
1.28665 

2.76330 
2.76766 
2.77200 

5.95334 
5.96273 
5.97209 

2.14 
2.15 
2.16 

4.5796 
4.6225 
4.6656 

1.46287 
1.46629 
1.46969 

4.62601 
4.63681 
4.64758 

9.80034 
9.93838 
10.0777 

1.28866 
1.29066 
1.29266 

2.77633 
2.78065 
2.78495 

5.98142 
5.99073 
6.00000 

2.17 
2.18 
2.19 

4.7089 
4.7524 
4.7961 

1.47309 
1.47648 
1.47986 

4.65833 
4.66905 
4.67974 

10.2183 
10.3602 
10.5035 

1.29465 
1.29664 
1.29862 

2.78924 
2.79352 
2.79779 

6.00925 
6.01846 
6.02765 

2.20 

4.8400 

1.48324 

4.69042 

10.6480 

1.30059 

2.80204 

6.03681 

2.21 
2.22 
2.23 

4.8841 
4.9284 
4.9729 

1.48661 
1.48997 
1.49332 

4.70106 
4.71169 
4.72229 

10.7939 
10.9410 
11.0896 

1.30256 
1.30452 
1.30648 

2.80628 
2.81050 
2.81472 

6.04594 
6.05505 
6.06413 

2.24 
2.25 
2.26 

5.0176 
5.0625 
5.1076 

1.49666 
1.50000 
1.50333 

4.73286 
4.74342 
4.75395 

11.2394 
11.3906 
11.5432 

1.30843 
1.31037 
1.31231 

2.81892 
2.82311 
2.82728 

6.07318 
6.08220 
6.09120 

2.27 

2.28 
2.29 

5.1529 
5.1984 
5.2441 

1.50665 
1.50997 
1.51327 

4.76445 
4.77493 
4.78539 

11.6971 
11.8524 
12.0090 

1.31424 
1.31617 
1.31809 

2.83145 
2.83560 
2.83974 

6.10017 
6.10911 
6.11803 

2.30 

5.2900 

1.51658 

4.79583 

12.1670 

1.32001 

2.84387 

6.12693 

2.31 
2.32 
2.33 

5.33G1 
5.3824 
5.4289 

1.51987 
1.52315 
1.52643 

4.80625 
4.81664 
4.82701 

12.3264 

12.4872 
12.6493 

1.32192 
1.32382 
1.32572 

2.84798 
2.85209 
2.85618 

6.13579 
6.14463 
6.15345 

2.34 
2.35 
2.36 

5.4756 
5.5225 
5.5696 

1.52971 
1.53297 
1.53623 

4.83735 
4.84768 
4.85798 

12.8129 
12.9779 
13.1443 

1.32761 
1.32950 
1.33139 

2.86026 
2.86433 
2.86838 

6.16224 
6.17101 
6.17975 

2.37 
2.38 
2.39 

5.6169 
5.6644 
5.7121 

1.53948 
1.54272 
1.54596 

4.86826 
4.87852 
4.88876 

13.3121 
13.4813 
13.6519 

1.33326 
1.33514 
1.33700 

2.87243 
2.87646 
2.88049 

6.18846 
6.19715 
6.20582 

2.40 

5.7600 

1.54919 

4.89898 

13.8240 

1.33887 

2.88450 

6.21447 

2.41 
2.42 
2.43 

5.8081 
5.8564 
5.9049 

1.55242 
1.55563 
1.55885 

4.90918 
4.91935 
4.92950 

13.9975 
14.1725 
14.3489 

1.34072 
1.34257 
1.34442 

2.88850 
2.89249 
2.89647 

6.22308 
6.23168 
6.24025 

2.44 
2.45 
2.46 

5.9536 
6.0025 
6.0516 

1.56205 
1.56525 
1.56844 

4.93964 
4.94975 
4.95984 

14.5268 
14.7061 
14.8869 

.34626 
.34810 
1.34993 

2.90044 
2.90439 
2.90834 

6.24880 
6.25732 
6.26583 

2.47 
2.48 
2.49 

6.1009 
6.1504 
6.2001 

1.57162 
1.57480 
1.57797 

4.96991 
4.97996 
4.98999 

15.0692 
15.2530 
15.4:382 

1.35176 
1.35358 
1.35540 

2.91227 
2.91620 
2.92011 

6.27431 
6.28276 
6.29119 

Powers  and  Roots 


281 


n 

W2 

Vn 

VlOw 

ns 

Vn 

VlOn 

^ioo^ 

2.50 

6.2500 

1.58114 

5.00000 

15.6250 

1.35721 

2.92402 

6.29961 

2.51 
2.52 
2.53 

6.3001 
6.3504 
6.4009 

1.58430 
1.58745 
1.59060 

5.00999 
5.01996 
5.02991 

15.8133 
16.0030 
16.1943 

1.35902 
1.36082 
1.36262 

2.92791 
2.93179 
2.93567 

6.30799 
6.31636 
6.32470 

2.54 
2.55 
2.56 

6.4516 
6.5025 
6.5536 

1.59374 
1.59687 
1.60000 

5.03984 
5.04975 
5.05964 

16.3871 
16.5814 
16.7772 

1.36441 
1.36620 
1.36798 

2.93953 
2.94338 
2.94723 

6.33303 
6.34133 
6.34960 

2.57 
2.58 
2.59 

6.6049 
6.6564 
6.7081 

1.60312 
1.60624 
1.60935 

5.06952 
5.07937 
5.08920 

16.9746 
17.1735 
17.3740 

1.36976 
1.37153 
1.37330 

2.95106 

2.95488 
2.95869 

6.35786 
6.36610 
6.37431 

2.60 

6.7600 

1.61245 

5.09902 

17.5760 

1.37507 

2.96250 

6.38250 

2.61 
2.62 
2.63 

6.8121 
6.8644 
6.9169 

1.61555 
1.61864 
1.62173 

5.10882 
5.11859 
5.12835 

17.7796 
17.9847 
18.1914 

1.37683 
1.37859 
1.38034 

2.96629 
2.97007 
2.97385 

6.39068 
6.39883 
6.40696 

2.64 
2.65 
2.66 

6.9696 
7.0225 
7.0756 

1.62481 

1.62788 
1.63095 

5.13809 
5.14782 
5.15752 

18.3997 
18.6096 
18.8211 

1.38208 
1.38383 
1.38557 

2.97761 
2.98137 
2.98511 

6.41507 
6.42316 
6.43123 

2.67 
2.68 
2.69 

7.1289 
7.1824 
7.2361 

1.63401 
1.63707 
1.64012 

5.16720 
5.17687 
5.18652 

19.0342 
19.2488 
19.4651 

1.38730 
1.38903 
1.39076 

2.98885 
2.99257 
2.99629 

6.43928 
6.44731 
6.45531 

2.70 

7.2900 

1.64317 

5.19615 

19.6830 

1.39248 

3.00000 

6.46330 

2.71 
2.72 
2.73 

7.3441 
7.3984 
7.4529 

1.64621 
1.64924 
1.65227 

5.20577 
5.21536 
5.22494 

19.9025 
20.1236 
20.3464 

1.39419 
1.39591 
1.39761 

3.00370 
3.00739 
3.01107 

6.47127 
6.47922 
6.48715 

2.74 
2.75 
2.76 

7.5076 
7.5625 
7.6176 

1.65529 
1.65831 
1.66132 

5.23450 
5.24404 
5.25357 

20.5708 
20.7969 
21.0246 

1.39932 
1.40102 
1.40272 

3.01474 
3.01841 
3.02206 

6.49507 
6.50296 
6.51083 

2.77 
2.78 
2.79 

7.6729 

7.7284 
7.7841 

1.66433 
1.66733 
1.67033 

5.26308 
5.27257 
5.28205 

21.2539 
21.4850 
21.7176 

1.40441 
1.40610 
1.40778 

3.02570 
3.02934 
3.03297 

6.51868 
6.52(552 
6.53434 

2.80 

7.8400 

1.67332 

5.29150 

21.9520 

1.40946 

3.03659 

6.54213 

2.81 
2.82 
2.83 

7.89G1 
7.9524 
8.0089 

1.67631 
1.67929 
1.68226 

5.30094 
5.31037 
5.31977 

22.1880 
22.4258 
22.6652 

1.41114 
1.41281 
1.41448 

3.04020 
3.04380 
3.04740 

6.54991 
6.55767 
6.56541 

2.84 
2.85 
2.86 

8.0656 
8.1225 
8.1796 

1.68523 
1.68819 
1.69115 

5.32917 
5.33854 
5.34790 

22.9063 
23.1491 
23.3937 

1.41614 
1.41780 
1.41946 

3.05098 
3.05456 
3.05813 

6.57314 
6.58084 
6.58853 

2.87 
2.88 
2.89 

8.2369 
8.2944 
8.3521 

1.69411 
1.69706 
1.70000 

5.35724 
5.36656 
5.37587 

23.6399 
23.8879 
24.1376 

1.42111 
1.42276 
1.42440 

3.06169 
3.06524 
3.06878 

6.59620 
6.60385 
6.61149 

2.90 

8.4100 

1.70294 

5.38516 

24.3890 

1.42604 

3.07232 

6.61911 

2.91 
2.92 
2.93 

8.4681 
8.5264 
8.5849 

1.70587 
1.70880 
1.71172 

5.39444 
5.40370 
5.41295 

24.6422 
24.8971 
25.1538 

1.42768 
1.42931 
1.43094 

3.07584 
3.07936 
3.08287 

6.62671 
6.63429 
6.64185 

2.94 
2.95 
2.96 

8.6436 
8.7025 
8.7616 

1.71464 
1.71756 
1.72047 

5.42218 
5.43139 
5.44059 

25.4122 
25.6724 
25.9343 

1.43257 
1.43*19 
1.43581 

3.08638 
3.08987 
3.09336 

6.64940 
6.65693 
6.66444 

2.97 
2.98 
2.99 

8.8209 
8.8804 
8.9401 

1.72337 
1.72627 
1.72916 

5.44977 
5.45894 
5.46809 

26.1981 
26.4636 
26.7309 

1.43743 
1.43904 
1.44065 

3.09684 
3.10031 
3.10378 

6.67194 
6.67942 
6.68688 

282 


Powers  and  Roots 


ft 

n2 

V^ 

VWn 

n* 

Vn 

VWn 

VlMn 

3.00 

9.0000 

1.73205 

5.47723 

27.0000 

1.44225 

3.10723 

6.69433 

3.01 
3.02 
3.03 

9.0601 
9.1204 
9.1809 

1.73494 
1.73781 
1.74069 

5.486:35 
5.49545 
5.50454 

27.2709 
27.5436 
27.8181 

1.44385 
1.44545 
1.44704 

3.11068 
3.11412 
3.11756 

6.70176 
6.70917 
6.71657 

3.04 
3.05 
3.06 

9.2416 
9.3025 
9.3636 

1.74356 
1.74642 
1.74929 

5.51362 
5.52268 
5.53173 

28.0945 
28.3726 
28.6526 

1.44863 
1.45022 
1.45180 

3.12098 
3.12440 
3.12781 

6.72395 
6.73132 
6.73866 

3.07 
3.08 
3.09 

9.4249 
9.4864 
9.5481 

1.75214 
1.75499 
1.75784 

5.54076 
5.54977 
5.55878 

28.9344 
29.2181 
29.5036 

1.45338 
1.45496 
1.45653 

3.13121 
3.13461 
3.13800 

6.74600 
6.75331 
6.76061 

3.10 

9.6100 

1.76068 

5.56776 

29.7910 

1.45810 

3.14138 

6.76790 

3.11 
3.12 
3.13 

9.6721 
9.7344 
9.7969 

1.76352 
1.76635 
1.76918 

5.57674 
5.58570 
5.59464 

30.0802 
30.3713 
30.6643 

1.45967 
1.46123 
1.46279 

3.14475 
3.14812 
3.15148 

6.77517 
6.78242 
6.78966 

3.14 

3.15 
3.16 

9.8596 
9.9225 
9.9856 

1.77200 
1.77482 
1.77764 

5.60357 
5.61249 
5.62139 

30.9591 
31.2559 
31.5545 

1.46434 
1.46590 
1.46745 

3.15483 
3.15818 
3.16152 

6.79688 
6.80409 
6.81128 

3.17 
3.18 
3.19 

10.0489 
10.1124 
10.1761 

1.78045 
1.78326 
1.78606 

5.63028 
5.63915 
5.64801 

31.8550 
32.1574 
32.4618 

1.46899 
1.47054 
1.47208 

3.16485 
3.16817 
3.17149 

6.81846 
6.82562 
6.83277 

3.20 

10.2400 

1.78885 

5.65685 

32.7680 

1.47361 

3.17480 

6.83990 

3.21 
3.22 
3.23 

10.3041 
10.3684 
10.4329 

1.79165 
1.79444 
1.79722 

5.66569 
5.67450 
5.68331 

33.0762 
33.3862 
33.6983 

1.47515 
1.47668 
1.47820 

3.17811 
3.18140 
3.18469 

6.84702 
6.85412 
6.86121 

3.24 
3.25 
3.26 

10.4976 
10.5625 
10.6276 

1.80000 
1.80278 
1.80555 

5.69210 
5.70088 
5.70964 

34.0122 
34.3281 
34.6460 

1.47973 
1.48125 

1.48277 

3.18798 
3.19125 
3.19452 

6.86829 
6.87534 
6.88239 

3.27 
3.28 
3.29 

10.6929 
10.7584 
10.8241 

1.80831 
1.81108 
1.81384 

5.71839 
5.72713 
5.73585 

34.9658 
35.2876 
35.6113 

1.48428 
1.48579 
1.48730 

3.19778 
3.20104 
3.20429 

6.88942 
6.89643 
6.90344 

3.30 

10.8900 

1.81659 

5.74456 

35.9370 

1.48881 

3.20753 

6.91042 

3.31 
3.32 
3.33 

10.9561 
11.0224 
11.0889 

1.81934 
1.82209 
1.82483 

5.75326 
5.76194 
5.77062 

36.2647 
36.5944 
36.9260 

1.49031 
1.49181 
1.49330 

3.21077 
3.21400 
3.21722 

6.91740 
6.92436 
6.93130 

3.34 
3.35 
3.36 

11.1556 
11.2225 
11.2896 

1.82757 
1.83030 
1.83303 

5.77927 
5.78792 
5.79655 

37.2597 
37.5954 
37.9331 

1.49480 
1.49629 
1.49777 

3.22044 
3.22365 
3.22686 

6.93823 
6.94515 
6.95205 

3.37 
3.38 
3.39 

11.3569 
11.4244 
11.4921 

1.83576 
1.83848 
1.84120 

5.80517 
5.81378 
5.82237 

38.2728 
38.6145 
38.9582 

1.49926 
1.50074 
1.50222 

3.23006 
3.23325 
3.23643 

6.95894 
6.96582 
6.97268 

3.40 

11.5600 

1.84391 

5.83095 

39.3040 

1.50369 

3.23961 

6.97953 

3.41 
3.42 
3.43 

11.6281 
11.6964 
11.7649 

1.84662 
1.84932 
1.85203 

5.83952 
5.84808 
6.85662 

39.6518 
40.0017 
40.3536 

1.50517 
1.50664 
1.50810 

3.24278 
3.24595 
3.24911 

6.98637 
6.99319 
7.00000 

3.44 
3.45 
3.46 

11.8336 
11.9025 
11.9716 

1.85472 
1.8.7742 
1.86011 

5.86515 
5.87367 
5.88218 

40.7076 
41.0636 
41.4217 

1.50957 
1.51103 
1.51249 

3.25227 
3.25542 
3.25856 

7.00680 
7.01358 
7.02035 

3.47 
3.48 
3.49 

12.0409 
12.1104 
12.1801 

1.86279 
1.86548 
1.86815 

5.89067 
5.89915 
5.90762 

41.7819 
42.1442 
42.5085 

1.51394 
1.51540 
1.51685 

3.26169 
3.26482 
3.26795 

7.02711 
7.03385 
7.04058 

Powers  and  Roots 


283 


n 

n? 

Vn 

vio™ 

n* 

^n 

VWH 

^lOOw 

3.50 

12.2500 

1.87083 

5.91608 

42.8750 

1.51829 

3.27107 

7.04730 

3.51 
3.52 
3.53 

12.3201 
12.3904 
12.4609 

1.87350 
1.87617 
1.87883 

5.92453 
5.93296 
5.94138 

43.2436 
43.6142 
43.9870 

1.51974 
1.52118 
1.52262 

3.27418 
3.27729 
3.28039 

7.05400 
7.06070 
7.06738 

3.54 
3.55 
3.56 

12.5316 
12.6025 
12.6736 

1.88149 
1.88414 
1.88680 

5.94979 
5.95819 
5.96657 

44.3619 
44.7389 
45.1180 

1.52406 
1.52549 
1.52692 

3.28348 
3.28657 
3.28965 

7.07404 
7.08070 
7.08734 

3.57 
3.58 
3.59 

12.7449 
12.8164 

12.8881 

1.88944 
1.89209 
1.89473 

5.97495 
5.98331 
5.99166 

45.4993 
45.8827 
46.2683 

1.52835 
1.52978 
1.53120 

3.29273 
3.29580 
3.29887 

7.09397 
7.10059 
7.10719 

3.60 

12.9600 

1.89737 

6.00000 

46.6560 

1.53262 

3.30193 

7.11379 

3.61 
3.62 
3.63 

13.0321 
13.1044 
13.1769 

1.90000 
1.90263 
1.90526 

6.00833 
6.01664 
6.02495 

47.0459 
47.4379 
47.8321 

1.53404 
1.53545 
1.53686 

3.30498 
3.30803 
3.31107 

7.12037 
7.12694 
7.13349 

3.64 
3.65 
3.66 

13.2496 
13.3225 
13.3956 

1.90788 
1.91050 
1.91311 

6.03324 
6.04152 
6.04979 

48.2285 
48.6271 
49.0279 

1.53827 
1.53968 
1.54109 

3.31411 
3.31714 
3.32017 

7.14004 
7.14657 
7.15309 

3.67 
3.68 
3.69 

13.4689 
13.5424 
13.6161 

1.91572 
1.91833 
1.92094 

6.05805 
6.06630 
6.07454 

49.4309 
49.8360 
50.2434 

1.54249 
1.54389 
1.54529 

3.32319 
3.32621 
3.32922 

7.15960 
7.16610 
7.17258 

3.70 

13.6900 

1.92354 

6.08276 

50.6530 

1.54668 

3.33222 

7.17905 

3.71 
3.72 
3.73 

13.7641 
13.8384 
13.9129 

1.92614 
1.92873 
1.93132 

6.09098 
6.09918 
6.10737 

51.0648 
51.4788 
51.8951 

1.54807 
1.54946 
1.55085 

3.33522 
3.33822 
3.34120 

7.18552 
7.19197 
7.19840 

3.74 
3.75 
3.76 

13.9876 
14.0625 
14.1376 

1.93391 
1.93649 
1.93907 

6.11555 
6.12372 
6.13188 

52.3136 
52.7344 
53.1574 

1.55223 
1.55362 
1.55500 

3.34419 
3.34716 
3.35014 

7.20483 
7.21125 
7.21765 

3.77 

3.78 
3.79 

14.2129 

14.2884 
14.3641 

1.94165 
1.94422 
1.94679 

6.14003 
6.14817 
6.15630 

53.5826 
54.0102 
54.4399 

1.55637 
1.55775 
1.55912 

3.35310 
3.35607 
3.35902 

7.22405 
7.23043 
7.23680 

3.80 

14.4400 

1.94936 

6.16441 

54.8720 

1.56049 

3.36198 

7.24316 

3.81 
3.82 
3.83 

14.5161 
14.5924 
14.6689 

1.95192 
1.95448 
1.95704 

6.17252 
6.18061 
6.18870 

55.3063 
55.7430 
56.1819 

1.56186 
1.56322 
1.56459 

3.36492 
3.36786 
3.37080 

7.24950 
7.25584 
7.26217 

3.84 
3.85 
3.86 

14.7456 
14.8225 
14.8996 

1.95959 
1.96214 
1.96469 

6.19677 
6.20484 
6.21289 

56.6231 
57.0666 
57.5125 

1.56595 
1.56731 
1.56866 

3.37373 
3.37666 
3.37958 

7.26848 
7.27479 
7.28108 

3.87 
3.88 
3.89 

14.9769 
15.0544 
15.1321 

1.96723 
1.96977 
1.97231 

6.22093 
6.22896 
6.23699 

57.9606 
58.4111 
58.8639 

1.57001 
1.57137 
1.57271 

3.38249 
3.38540 
3.38831 

7.28736 
7.29363 
7.29989 

3.90 

15.2100 

1.97484 

6.24500 

59.3190 

1.57406 

3.39121 

7.30614 

3.91 
3.92 
3.93 

15.2881 
15.3664 
15.4449 

1.97737 
1.97990 
1.98242 

6.25300 
6.26099 
6.26897 

59.7765 
60.2363 
60.6985 

1.57541 
1.57675 
1.57809 

3.39411 
3.39700 
3.39988 

7.31238 
7.31861 
7.32483 

3.94 
3.95 
3.96 

15.5236 
15.6025 
15.6816 

1.98494 
1.98746 
1.98997 

6.27694 
6.28490 
6.29285 

61.1630 
61.6299 
62.0991 

1.57942 
1.58076 
1.58209 

3.40277 
3.40564 
3.40851 

7.33104 
7.33723 
7.34342 

3.97 
3.98 
3.99 

15.7609 
15.8404 
15.9201 

1.99249 
1.99499 
1.99750 

6.30079 
6.30872 
6.31664 

62.5708 
63.0448 
63.5212 

1.58342 
1.58475 
1.58608 

3.41138 
3.41424 
3.41710 

7.34960 
7.35576 
7.36192 

284 


Powers  and  Roots 


n 

n2 

V^ 

VWn 

n? 

Vn 

i/Wn 

VIMn 

4.00 

16.0000 

2.00000 

6.32456 

64.0000 

1.58740 

3.41995 

7.36806 

4.01 
4.02 
4.03 

16.0801 
16.1604 
16.2409 

2.00250 
2.00499 
2.00749 

6.33246 
6.34035 
6.34823 

64.4812 
64.9648 
65.4508 

1.58872 
1.59004 
1.59136 

3.42280 
3.42564 
3.42848 

7.37420 
7.38032 
7.38644 

4.04 
4.05 
4.06 

16.3216 
16.4025 
16.4836 

2.00998 
2.01246 
2.01494 

6.35610 
6.36396 
6.37181 

65.9393 
66.4301 
66.9234 

1.59267 
1.59399 
1.59530 

3.43131 
3.43414 
3.43697 

7.39254 
7.39864 
7.40472 

4.07 
4.08 
4.09 

16.5649 
16.6464 
16.7281 

2.01742 
2.01990 
2.02237 

6.37966 
6.38749 
6.39531 

67.4191 
67.9173 
68.4179 

1.59661 
1.59791 
1.59922 

3.43979 
3.44260 
3.44541 

7.41080 
7.41686 
7.42291 

4.10 

16.8100 

2.02485 

6.40312 

68.9210 

1.60052 

3.44822 

7.42896 

4.11 
4.12 
4.13 

16.8921 
16.9744 
17.0569 

2.02731 
2.02978 
2.03224 

6.41093 
6.41872 
6.42651 

69.4265 
69.9345 
70.4450 

1.60182 
1.60312 
1.60441 

3.45102 
3.45382 
3.45661 

7.43499 
7.44102 
7.44703 

4.14 
4.15 
4.16 

17.1396 
17.2225 
17.3056 

2.03470 
2.03715 
2.03961 

6.43428 
6.44205 
6.44981 

70.9579 
71.4734 
71.9913 

1.60571 
1.60700 
1.60829 

3.45939 
3.46218 
3.46496 

7.45304 
7.45904 
7.46502 

4.17 
4.18 
4.19 

17.3889 
17.4724 
17.5561 

2.04206 
2.04450 
2.04695 

6.45755 
6.46529 
6.47302 

72.5117 
73.0346 
73.5601 

1.60958 
1.61086 
1.61215 

3.46773 
3.47050 
3.47327 

7.47100 
7.47697 
7.48292 

4.20 

17.6400 

2.04939 

6.48074 

74.0880 

1.61343 

3.47603 

7.48887 

4.21 
4.22 
4.23 

17.7241 

17.8084 
17.8929 

2.05183 
2.05426 
2.05670 

6.48845 
6.49615 
6.50384 

74.6185 
75.1514 
75.6870 

1.61471 
1.61599 
1.61726 

3.47878 
3.48154 
3.48428 

7.49481 
7.50074 
7.50666 

4.24 
4.25 
4.26 

17.9776 
18.0625 
18.1476 

2.05913 
2.06155 
2.06398 

6.51153 
6.51920 
6.52687 

76.2250 
76.7656 
77.3088 

1.61853 
1.61981 
1.62108 

3.48703 
3.48977 
3.49250 

7.51257 
7.51847 
7.52437 

4.27 
4.28 
4.29 

18.2329 
18.3184 
18.4041 

2.06640 
2.06882 
2.07123 

6.53452 

6.54217 
6.54981 

77.8545 
78.4028 
78.9536 

1.62234 
1.62361 
1.62487 

3.49523 
3.49796 
3.50068 

7.53025 
7.53612 
7.54199 

4.30 

18.4900 

2.07364 

6.55744 

79.5070 

1.62613 

3.50340 

7.54784 

4.31 
4.32 
4.33 

18.5761 
18.6624 
18.7489 

2.07605 
2.07846 
2.08087 

6.56506 
6.57267 
6.58027 

80.0630 
80.6216 
81.1827 

1.62739 
1.62865 
1.62991 

3.50611 
3.50882 
3.51153 

7.55369 
7.55953 
7.56535 

4.34 
4.35 
4.36 

18.8356 
18.9225 
19.0096 

2.08327 
2.08567 
2.08806 

6.58787 
6.59545 
6.60303 

81.7465 
82.3129 
82.8819 

1.63116 
1.63241 
1.63366 

3.51423 
3.51692 
3.51962 

7.57117 
7.57698 
7.58279 

4.37 
4.38 
4.39 

19.0969 
19.1844 
19.2721 

2.09045 
2.09284 
2.09523 

6.61060 
6.61816 
6.62571 

83.4535 
84.0277 
84.6045 

1.63491 
1.63619 
1.63740 

3.52231 
3.52499 
3.52767 

7.58858 
7.59436 
7.60014 

4.40 

19.3600 

2.09762 

6.63325 

85.1840 

1.63864 

3.53035 

7.60590 

4.41 
4.42 
4.43 

19.4481 
19.5364 
19.6249 

2.10000 
2.10238 
2.10476 

6.64078 
6.64831 
6.65582 

85.7661 
86.3509 
86.9383 

1.63988 
1.64112 
1.64236 

3.53302 
3.53569 
3.53835 

7.61166 
7.61741 
7.62316 

4.44 
4.45 
4.46 

19.7136 
19.8025 
19.8916 

2.10713 
2.10950 
2.11187 

6.66333 
6.67083 
6.67832 

87.5284 
88.1211 
88.7165 

1.64359 
1.64483 
1.64606 

3.54101 
3.54367 
3.54632 

7.62888 
7.63461 
7.64032 

4.47 
4.48 
4.49 

19.9809 
20.0704 
20.1601 

2.11424 
2.11660 
2.11896 

6.68581 
6.69328 
6.70075 

89.3146 
89.9154 

90.5188 

1.64729 
1.64851 

1.64974 

3.54897 
3.55162 
3.55426 

7.64603 
7.65172 
7.65741 

Powers  and  Roots 


285 


n 

n2 

Vw 

VlOw 

n8 

Vn 

^10  n 

\/100w 

4.50 

20.2500 

2.12132 

6.70820 

91.1250 

1.65096 

3.55689 

7.66309 

4.51 
4.52 
4.53 

20.3401 
20.4304 
20.5209 

2.12368 
2.12603 
2.12838 

6.71565 
6.72309 
6.73053 

91.7339 
92.3454 
92.9597 

1.65219 
1.65341 
1.65462 

3.55953 
3.56215 
3.56478 

7.66877 
7.67443 
7,68009 

4.54 
4.55 
4.56 

20.6116 
20.7025 
20.7936 

2.13073 
2.13307 
2.13542 

6.73795 
6.74537 
6.75278 

93.5767 
94.1964 
94.8188 

1.65584 
1.65706 
1.65827 

3.56740 
3.57002 
3.57263 

7.68573 
7.69137 
7.69700 

4.57 
4.58 
4.59 

20.8849 
20.9764 
21.0681 

2.13776 
2.14009 
2.14243 

6.76018 
6.76757 
6.77495 

95.4440 
96.0719 
96.7026 

1.65948 
1.66069 
1.66190 

3.57524 
3.57785 
3.58045 

7.70262 
7.70824 
7.71384 

4.60 

21.1600 

2.14476 

6.78233 

97.3360 

1.66310 

3.58305 

7.71944 

4.61 
4.62 
4.63 

21.2521 
21.3444 
21.4369 

2.14709 
2.14942 
2.15174 

6.78970 
6.79706 
6.80441 

97.9722 
98.6111 
99.2528 

1.66431 
1.66551 
1.66671 

3.58564 
3.58823 
3.59082 

7.72503 
7.73061 
7.73619 

4.64 
4.65 
4.66 

21.5296 
21.6225 
21.7156 

2.15407 
2.15639 
2.15870 

6.81175 
6.81909 
6.82642 

99.8973 
100.545 
101.195 

1.66791 

1.66911 
1.67030 

3.59340 
3.59598 
3.59856 

7.74175 
7.74731 

7.75286 

4.67 
4.68 
4.69 

21.8089 
21.9024 
21.9961 

2.16102 
2.16333 
2.16564 

6.83374 
6.84105 
6.84836 

101.848 
102.503 
103.162 

1.67150 
1.67269 
1.67388 

3.60113 
3.60370 
3.60626 

7.75840 
7.76394 
7.76946 

4.70 

22.0900 

2.16795 

6.85565 

103.823 

1.67507 

3.60883 

7.77498 

4.7? 
4.72 
4.73 

22.1841 
22.2784 
22.3729 

2.17025 
2.17256 
2.17486 

6.86294 
6.87023 
6.87750 

104.487 
105.154 
105.824 

1.67626 
1.67744 
1.67863 

3.61138 
3.61394 
3.61649 

7.78049 
7.78599 
7.79149 

4.74 
4.75 
4.76 

22.4676 
22.5625 
22.6576 

2.17715 
2.17945 
2.18174 

6.88477 
6.89202 
6.89928 

106.496 
107.172 
107.850 

1.67981 
1.68099 
1.68217 

3.61903 
3.62158 
3.62412 

7.79697 
7.80245 
7.80793 

4.77 
4.78 
4.79 

22.7529 

22.8484 
22.9441 

2.18403 
2.18632 
2.18861 

6.90652 
6.91375 
6.92098 

108.531 
109.215 
109.902 

1.68334 
1.68452 
1.68569 

3.62665 
3.62919 
3.63172 

7.81339 
7.81885 
7.82429 

4.80 

23.0400 

2.19089 

6.92820 

110.592 

1.68687 

3.63424 

7.82974 

4.81 
4.82 
4.83 

23.1361 
23.2324 
23.3289 

2.19317 
2.19545 
2.19773 

6.93542 
6.94262 
6.94982 

111.285 
111.980 
112.679 

1.68804 
1.68920 
1.69037 

3.63676 
3.63928 
3.64180 

7.83517 
7.84059 
7.84601 

4.84 
4.85 
4.86 

23.4256 
23.5225 
23.6196 

2.20000 
2.20227 
2.20454 

6.95701 
6.96419 
6.97137 

113.380 
114.084 
114.791 

1.69154 
1.69270 
1.69386 

3.64431 
3.64682 
3.64932 

7.85142 
7.85683 
7.86222 

4.87 
4.88 
4.89 

23.7169 
23.8144 
23.9121 

2.20681 
2.20907 
2.21133 

6.97854 
6.98570 
6.99285 

115.501 
116.214 
116.930 

1.69503 
1.69619 
1.69734 

3.65182 
3.65432 
3.65681 

7.86761 
7.87299 
7.87837 

4.90 

24.0100 

2.21359 

7.00000 

117.649 

1.69850 

3.65931 

7.88374 

4.91 
4.92 
4.93 

24.1081  • 
24.2064 
24.3049 

2.21585 
2.21811 
2.22036 

7.00714 
7.01427 
7.02140 

118.371 
119.095 
119.823 

1.69965 
1.70081 
1.70196 

3.66179 
3.66428 
3.66676 

7.88909 
7.89445 
7.89979 

4.94 
4.95 
4.96 

24.4036 
24.5025 
24.6016 

2.22261 
2.22486 
2.22711 

7.02851 
7.03562 
7.04273 

120.554 
121.287 
122.024 

1.70311 
1.70426 
1.70540 

3.66924 
3.67171 
3.67418 

7.90513 
7.91046 
7.91578 

4.97 
4.98 
4.99 

24.7009 
24.8004 
24.9001 

2.22935 
2.23159 

2.23383 

7.04982 
7.05691 
7.06399 

122.763 
123.506 
124.251 

1.70655 
1.70769 

1.70884 

3.67665 
3.67911 
3.68157 

7.92110 
7.92641 
7.93171 

286 


Powers  and  Roots 


n 

W2 

Vn 

VlOw 

W8 

Vn 

vwn 

^iooli 

5.00 

25.0000 

2.23607 

7.07107 

125.000 

1.70998 

3.68403 

7.93701 

5.01 
5.02 
5.03 

25.1001 
25.2004 
25.3009 

2.23830 
2.24054 
2.24277 

7.07814 
7.08520 
7.09225 

125.752 
126.506 
127.264 

1.71112 
1.71225 
1.71339 

3.68649 
3.68894 
3.69138 

7.94229 
7.94757 
7.95285 

5.04 
5.05 
5.06 

25.4016 
25.5025 
25.6036 

2.24499 
2.24722 
2.24944 

7.09930 
7.10634 
7.11337 

128.024 

128.788 
129.554 

1.71452 
1.71566 
1.71679 

3.69383 
3.69627 
3.69871 

7.95811 
7.96337 
7.96863 

5.07 
5.08 
5.09 

25.7049 
25.8064 
25.9081 

2.25167 
2.25389 
2.25610 

7.12039 
7.12741 
7.13442 

130.324 
131.097 
131.872 

1.71792 
1.71905 
1.72017 

3.70114 
3.70357 
3.70600 

7.97387 
7.97911 
7.98434 

5.10 

26.0100 

2.25832 

7.14143 

132.651 

1.72130 

3.70843 

7.98957 

5.11 
5.12 
5.13 

26.1121 
26.2144 
26.3169 

2.26053 
2.20274 
2.26495 

7.14843 
7.15542 
7.16240 

133.433 
134.218 
135.006 

1.72242 
1,72355 
1.72467 

3.71085 
3.71327 
3.71569 

7.99479 
8.00000 
8.00520 

5.14 
5.15 
5.16 

26.4196 
26.5225 
26.6256 

2.26716 
2.26936 
2.27156 

7.16938 
7.17635 
7.18331 

135.797 
136.591 
137.388 

1.72579 
1.72691 

1.72802 

3.71810 
3.72051 
3.72292 

8.01040 
8.01559 
8.02078 

5.1T 
5.18 
5.19 

26.7289 
26.8324 
26.9361 

2.27376 
2.27596 
2.27816 

7.19027 
7.19722 
7.20417 

138.188 
138.992 
139.798 

1.72914 
1.73025 
1.73137 

3.72532 
3,72772 
3.73012 

8.02596 
8.03113 
8.03629 

5.20 

27.0400 

2.28035 

7.21110 

140.608 

1.73248 

3.73251 

8.04145 

5.21 
5.22 
5.23 

27.1441 
27.2484 
27.3529 

2.28254 
2.28473 
2.28692 

7.21803 
7.22496 
7.23187 

141.421 
142.237 
143.056 

1.73359 
1.73470 
1.73580 

3.73490 
3.73729 
3.73968 

8.04660 
8.05175 
8.05689 

5.24 
5.25 
5.26 

27.4576 
27.5625 
27.6676 

2.28910 
2.29129 
2.29347 

7.23878 
7.24569 
7.25259 

143.878 
144.703 
145.532 

1.73691 

1.73801 
1.73912 

3.74206 
3.74443 
3.74681 

8.06202 
8.06714 
8.07226 

5.27 
5.28 
5.29 

27.7729 
27.8784 
27.9841 

2.29565 
2.29783 
2.30000 

7.25948 
7.26636 
7.27324 

146.363 
147.198 
148.036 

1.74022 
1.74132 
1.74242 

3.74918 
3.75155 
3.75392 

8.07737 
8.08248 
8.08758 

5.30 

28.0900 

2.30217 

7.28011 

148.877 

1.74351 

3.75629 

8.09267 

5.31 
5.32 
5.33 

28.1961 
28.3024 
28.4089 

2.30434 
2.30651 
2.30868 

7.28697 
7.29383 
7.30068 

149.721 
150.569 
151.419 

1.74461 
1.74570 
1.74680 

3.75865 
3.76101 
3.76336 

8.09776 
8.10284 
8.10791 

5.34 
5.35 
5.36 

28.5156 
28.6225 
28.7296 

2.31084 
2.31301 
2.31517 

7.30753 
7.31437 
7.32120 

152.273 
153.130 
153.991 

1.74789 
1.74898 
1.75007 

3.76571 
3.76806 
3.77041 

8.11298 
8.11804 
8.12310 

5.37 
5.38 
5.39 

28.8369 
28.9444 
29.0521 

2.31733 
2.31948 
2.32164 

7.32803 
7.33485 
7.34166 

154.854 
155.721 
156.591 

1.75116 
1.75224 
1.75333 

3.77275 
3.77509 
3.77743 

8.12814 
8.13319 
8.13822 

5.40 

29.1600 

2.32379 

7.34847 

157.464 

1.75441 

3.77976 

8.14325 

5.41 
5.42 
5.43 

29.2681 
29.3764 
29.4849 

2.32594 
2.32809 
2.33024 

7.35527 
7.36206 
7.36885 

158.340 
159.220 
160.103 

1.75549 
1.75657 
1.75765 

3.78209 
3.78442 
3.78675 

8.14828 
8.15329 
8.15831 

5.44 
5.45 
5.46 

29.5936 
29.7025 
29.8116 

2.33238 
2.33452 
2.33666 

7.37564 
7.38241 
7.38918 

160.989 
161.879 
162.771 

1.75873 
1.75981 
1.76088 

3.78907 
3.79139 
3.79371 

8.16331 
8.16831 
8.17330 

5.47 
5.48 
5.49 

29.9209 
30.0304 
30.1401 

2.33880 
2.34094 
2.34307 

7.39594 
7.40270 
7.40945 

163.667 
164.567 
165.469 

1.76196 
1.76303 
1.76410 

3.79603 
3.79834 
3.80065 

8.17829 
8.18327 
8.18824 

Powers  and  Roots 


287 


ft 

n* 

•Vn 

vio™ 

ns 

#n 

VWn 

^100^ 

5.50 

30.2500 

2.34521 

7.41620 

166.375 

1.76517 

3.80295 

8.19321 

5.51 
5.52 
5.53 

30.3601 
30.4704 
30.5809 

2.34734 
2.34947 
2.35160 

7.42294 
7.42967 
7.43640 

167.284 
168.197 
169.112 

1.76624 
1.76731 
1.76838 

3.80526 
3.80756 
3.80985 

8.19818 
8.20313 
8.20808 

5.54 
5.55 
5.56 

30.6916 
30.8025 
30.9136 

2.35372 
2.35584 
2.35797 

7.44312 
7.44983 
7.45654 

170.031 
170.954 
171.880 

1.76944 
1.77051 
1.77157 

3.81215 
3.81444 
3.81673 

8.21303 

8.21797 
8.22290 

5.57 
5.58 
5.59 

31.0249 
31.1364 
31.2481 

2.36008 
2.36220 
2.36432 

7.46324 
7.46994 
7.47663 

172.809 
173.741 
174.677 

1.77263 
1.77369 
1.77475 

3.81902 
3.82130 
3.82358 

8.22783 
8.23275 
8.23766 

5.60 

31.3600 

2.36643 

7.48331 

175.616 

1.77581 

3.82586 

8.24257 

5.61 
5.62 
5.63 

31.4721 
31.5844 
31.6969 

2.36854 
2.37065 
2.37276 

7.48999 
7.49667 
7.50333 

176.558 
177.504 

178.454 

1.77686 
1.77792 
1.77897 

3.82814 
3.83041 
3.83268 

8.24747 
8.25237 
8.25726 

5.64 
5.65 
5.66 

31.8096 
31.9225 
32.0356 

2.37487 
2.37697 
2.37908 

7.50999 
7.51665 
7.52330 

179.406 
180.362 
181.321 

1.78003 
1.78108 
1.78213 

3.83495 
3.83722 
3.83948 

8.26215 
8.26703 
8.27190 

5.67 
5.68 
5.69 

32.1489 
32.2624 
32.3761 

2.38118 
2.38328 
2.38537 

7.52994 
7.53658 
7.54321 

182.284 
183.250 
184.220 

1.78318 
1.78422 
1.78527 

3.84174 
3.84399 
3.84625 

8.27677 
8.28164 
8.28649 

5.70 

32.4900 

2.38747 

7.54983 

185.193 

1.78632 

3.84850 

8.29134 

5.71 
5.72 
5.73 

32.6041 
32.7184 
32.8329 

2.38956 
2.39165 
2.39374 

7.55645 
7.56307 
7.56968 

186.169 
187.149 
188.133 

1.78736 
1.78840 
1.78944 

3.85075 
3.85300 
3.85524 

8.29619 
8.30103 
8.30587 

5.74 
5.75 
5.76 

32.9476 
33.0625 
33.1776 

2.39583 
2.39792 
2.40000 

7.57628 
7.58288 
7.58947 

189.119 
190.109 
191.103 

1.79048 
1.79152 
1.79256 

3.85748 
3.85972 
3.86196 

8.31069 
8.31552 
8.32034 

5.77 
5.78 
5.79 

33.2929 
33.4084 
33.5241 

2.40208 
2.40416 
2.40624 

7.59605 
7.60263 
7.60920 

192.100 
193.101 
194.105 

1.79360 
1.79463 
1.79567 

3.86419 
3.86642 
3.86865 

8.32515 
8.32995 
8.33476 

5.80 

33.6400 

2.40832 

7.61577 

195.112 

1.79670 

3.87088 

8.33955 

5.81 

5.82 
5.83 

33.7561 
33.8724 
33.9889 

2.41039 
2.41247 
2.41454 

7.62234 
7.62889 
7.63544 

196.123 
197.137 
198.155 

1.79773 
1.79876 
1.79979 

3.87310 
3.87532 
3.87754 

8.34434 
8.34913 
8.35390 

5.84 
5.85 
5.86 

34.1056 
34.2225 
34.3396 

2.41661 
2.41868 
2.42074 

7.64199 
7.64853 
7.65506 

199.177 
200.202 
201.230 

1.80082 
1.80185 
1.80288 

3.87975 
3.88197 
3.88418 

8.35868 
8.36345 
8.36821 

5.87 
5.88 
5.89 

34.4569 
34.5744 
34.6921 

2.42281 
2.42487 
2.42693 

7.66159 
7.66812 
7.67463 

202.262 
203.297 
204.336 

1.80390 
1.80492 
1.80595 

3.88639 
3.88859 
3.89080 

8.37297 

8.37772 
8.38247 

5.90 

34.8100 

2.42899 

7.68115 

205.379 

1.80697 

3.89300 

8.38721 

5.91 
5.92 
5.93 

34.9281 
35.0464 
35.1649 

2.43105 
2.43311 
2.43516 

7.68765 
7.69415 
7.70065 

206.425 
207.475 
208.528 

1.80799 
1.80901 
1.81003 

3.89519 
3.89739 
3.89958 

8.39194 
8.39667 
8.40140 

5.94 
5.95 
5.96 

35.2836 
35.4025 
35.5216 

2.43721 
2.48926 
2.44131 

7.70714 
7.71362 
7.72010 

209.585 
210.645 
211.709 

1.81104 
1.81206 
1.81307 

3.90177 
3.90396 
3.90615 

8.40612 
8.41083 
8.41554 

5.97 
5.98 
5.99 

35.6409 
35.7604 
35.8801 

2.44336 
2.44540 

2.44745 

7.72658 
7.73305 
7.73951 

212.776 
213.847 
214.922 

1.81409 
1.81510 

1.81611 

3.90833 
3.91051 
3.91269 

8.42025 
8.42494 
8.42964 

288 


Powers  and  Boots 


n 

w2 

Vw 

VWJi 

W3 

va 

VWn 

^100™ 

6.00 

36.0000 

2.44949 

7.74597 

216.000 

1.81712 

3.91487 

8.43433 

6.01 

36.1201 

2.45153 

7.75242 

217.082 

1.81813 

3.91704 

8.43901 

6.02 

36.2404 

2.45357 

7.75887 

218.167 

1.81914 

3.91921 

8.44369 

6.03 

36.3609 

2.45561 

7.76531 

219.256 

1.82014 

3.92138 

8.44836 

6.04 

36.4816 

2.45764 

7.77174 

220.349 

1.82115 

3.92355 

8.45303 

6.05 

86.6025 

2.45967 

7.77817 

221.445 

1.82215 

3.92571 

8.45769 

6.06 

36.7236 

2.46171 

7.78460 

222.545 

1.82316 

3.92787 

8.46235 

6.07 

36.8449 

2.46374 

7.79102 

223.649 

1.82416 

3.93003 

8.46700 

6.08 

36.9664 

2.46577 

7.79744 

224.756 

1.82516 

3.93219 

8.47165 

6.09 

37.0881 

2.46779 

7.80385 

225.867 

1.82616 

3.93434 

8.47629 

6.10 

37.2100 

2.46982 

7.81025 

226.981 

1.82716 

3.93650 

8.48093 

6.11 

37.3321 

2.47184 

7.81665 

228.099 

1.82816 

3.93865 

8.48556 

6.12 

37.4544 

2.47386 

7.82304 

229.221 

1.82915 

3.94079 

8.49018 

6.13 

37.5769 

2.47588 

7.82943 

230.346 

1.83015 

3.94294 

8.49481 

6.14 

37.6996 

2.47790 

7.83582 

231.476 

1.83115 

3.94508 

8.49942 

6.15 

37.8225 

2.47992 

7.84219 

232.608 

1.83214 

3.94722 

8.50403 

6.16 

37.9456 

2.48193 

7.84857 

233.745 

1.83313 

3.94936 

8.50864 

6.17 

38.0689 

2.48395 

7.85493 

234.885 

1.83412 

3.95150 

8.51324 

6.18 

38.1924 

2.48596 

7.86130 

236.029 

1.83511 

3.95363 

8.51784 

6.19 

38.3161 

2.48797 

7.86766 

237.177 

1.83610 

3.95576 

8.52243 

6.20 

38.4400 

2.48998 

7.87401 

238.328 

1.83709 

3.95789 

8.52702 

6.21 

38.5641 

2.49199 

7.88036 

239.483 

1.83808 

3.96002 

8.53160 

6.22 

38.6884 

2.49399 

7.88670 

240.642 

1.83906 

3.96214 

8.53618 

6.23 

38.8129 

2.49600 

7.89303 

241.804 

1.84005 

3.96427 

8.54075 

6.24 

38.9376 

2.49800 

7.89937 

242.971 

1.84103 

3.96638 

8.54532 

6.25 

39.0625 

2.50000 

7.90569 

244.141 

1.84202 

3.96850 

8.54988 

6.26 

39.1876 

2.50200 

7.91202 

245.314 

1.84300 

3.97062 

8.55444 

6.27 

39.3129 

2.50400 

7.91833 

246.492 

1.84398 

3.97273 

8.55899 

6.28 

39.4384 

2.50599 

7.92465 

247.673 

1.84496 

3.97484 

8.56354 

6.29 

39.5641 

2.50799 

7.93095 

248.858 

1.84594 

3.97695 

8.56808 

6.30 

39.6900 

2.50998 

7.93725 

250.047 

1.84691 

3.97906 

8.57262 

6.31 

39.8161 

2.51197 

7.94355 

251.240 

1.84789 

3.98116 

8.57715 

6.32 

39.9424 

2.51396 

7.94984 

252.436 

1.84887 

3.98326 

8.58168 

6.33 

40.0689 

2.51595 

7.95613 

253.636 

1.84984 

3.98536 

8.58620 

6.34 

40.1956 

2.51794 

7.96241 

254.840 

1.85082 

3.98746 

8.59072 

6.35 

40.3225 

2.51992 

7.96869 

256.048 

1.85179 

3.98956 

8.59524 

6.36 

40.4496 

2.52190 

7.97496 

257.259 

1.85276 

3.99165 

8.59975 

6.37 

40.5769 

2.52389 

7.98123 

258.475 

1.85373 

3.99374 

8.60425 

6.38 

40.7044 

2.52587 

7.98749 

259.694 

1.85470 

3.99583 

8.60875 

6.39 

40.8321 

2.52784 

7.99375 

260.917 

1.85567 

3.99792 

8.61325 

6.40 

40.9600 

2.52982 

8.00000 

262.144 

1.85664 

4.00000 

8.61774 

6.41 

41.0881 

2.53180 

8.00625 

263.375 

1.85760 

4.00208 

8.62222 

6.42 

41.2164 

2.53377 

8.01249 

264.609 

1.85857 

4.00416 

8.<»2671 

6.43 

41.3449 

2.53574 

8.01873 

265.848 

1.85953 

4.00624 

8.63118 

6.44 

41.4736 

2.53772 

8.02496 

267.090 

1.86050 

4.00832 

8.63566 

6.45 

41.6025 

2.53969 

8.03119 

268.336 

1.86146 

4.01039 

8.64012 

6.46 

41.7316 

2.54165 

8.03741 

269.586 

1.86242 

4.01246 

8.64459 

6.47 

41.8609 

2.54362 

8.04363 

270.840 

1.86338 

4.01453 

8.64904 

6.48 

41.9904 

2.r>4.r>r,,s 

8.04984 

272.098 

1.86434 

4.01660 

8.  65350 

6.49 

42.1201 

2.54755 

8.05605 

273.359 

1.8651*0 

4.01866 

s.<;r,7<)5 

Powers  and  Roots 


289 


n 

n2 

^n 

vio™ 

n8 

Vn 

vlO  n 

^100^ 

6.50 

42.2500 

2.54951 

8.06226 

274.625 

1.86626 

4.02073 

8.66239 

6.51 
6.52 
6.53 

42.3801 
42.5104 
42.6409 

2.55147 
2.55343 
2.55539 

8.06846 
8.07465 
8.08084 

275.894 
277.168 
278.445 

1.86721 
1.86817 
1.86912 

4.02279 
4.02485 
4.02690 

8.66683 
8.67127 
8.67570 

6.54 
6.55 
6.56 

42.7716 
42.9025 
43.0336 

2.55734 
2.55930 
2.56125 

8.08703 
8.09321 
8.09938 

279.726 
281.011 
282.300 

1.87008 
1.87103 
1.87198 

4.02896 
4.03101 
4.03306 

8.68012 
8.68455 
8.68896 

6.57 
6.58 
6.59 

43.1649 
43.2964 
43.4281 

2.56320 
2.56515 
2.56710 

8.10555 
8.11172 
8.11788 

283.593 
284.890 
286.191 

1.87293 
1.87388 
1.87483 

4.03511 
4.03715 
4.03920 

8.69338 
8.69778 
8.70219 

6.60 

43.5600 

2.56905 

8.12404 

287.496 

1.87578 

4.04124 

8.70659 

6.61 
6.62 
6.63 

43.6921 
43.8244 
43.9569 

2.57099 
2.57294 
2.57488 

8.13019 
8.13634 
8.14248 

288.805 
290.118 
291.434 

1.87672 
1.87767 
1.87862 

4.04328 
4.04532 
4.04735 

8.71098 
8.71537 
8.71976 

6.64 
6.65 
6.66 

44.0896 
44.2225 
44.3556 

2.57682 
2.57876 
2.58070 

8.14862 
8.15475 
8.16088 

292.755 
294.080 
295.408 

1.87956 
1.88050 
1.88144 

4.04939 
4.05142 
4.05345 

8.72414 
8.72852 
8.73289 

6.67 
6.68 
6.69 

44.4889 
44.6224 
44.7561 

2.58263 
2.58457 
2.58650 

8.16701 
8.17313 
8.17924 

296.741 
298.078 
299.418 

1.88239 
1.88333 
1.88427 

4.05548 
4.05750 
4.05953 

8.73726 
8.74162 
8.74598 

6.70 

44.8900 

2.58844 

8.18535 

300.763 

1.88520 

4.06155 

8.75034 

6.71 
6.72 
6.73 

45.0241 
45.1584 
45.2929 

2.59037 
2.59230 
2.59422 

8.19146 
8.19756 
8.20366 

302.112 
303.464 
304.821 

1.88614 
1.88708 
1.88801 

4.06357 
4.06559 
4.06760 

8.75469 
8.75904 
8.76338 

6.74 
6.75 
6.76 

45.4276 
45.5625 
45.6976 

2.59615 
2.59808 
2.60000 

8.20975 
8.21584 
8.22192 

306.182 
307.547 
308.916 

1.88895 
1.88988 
1.89081 

4.06961 
4.07163 
4.07364 

8.76772 
8.77205 
8.77638 

6.77 
6.78 
6.79 

45.8329 
45.9684 
46.1041 

2.60192 
2.60384 
2.60576 

8.22800 
8.23408 
8.24015 

310.289 
311.666 
313.047 

1.89175 
1.89268 
1.89361 

4.07564 
4.07765 
4.07965 

8.78071 
8.78503 
8.78935 

6.80 

46.2400 

2.60768 

8.24621 

314.432 

1.89454 

4.08166 

8.79366 

6.81 
6.82 
6.83 

46.3761 
46.5124 
46.6489 

2.60960 
2.61151 
2.61343 

8.25227 
8.25833 
8.26438 

315.821 
317.215 
318.612 

1.89546 
1.89639 
1.89732 

4.08365 
4.08565 
4.08765 

8.79797 
8.80227 
8.80657 

6.84 
6.85 
6.86 

46.7856 
46.9225 
47.0596 

2.61534 
2.61725 
2.61916 

8.27043 
8.27647 
8.28251 

320.014 
321.419 
322.829 

1.89824 
1.89917 
1.90009 

4.08964 
4.09163 
4.09362 

8.81087 
8.81516 
8.81945 

6.87 
6.88 
6.89 

47.1969 
47.3344 
47.4721 

2.62107 
2.62298 
2.62488 

8.28855 
8.29458 
8.30060 

324.243 
325.661 
327.083 

1.90102 
1.90194 
1.90286 

4.09561 
4.09760 
4.09958 

8.82373 
8.82801 
8.83228 

6.90 

47.6100 

2.62679 

8.30662 

328.509 

1.90378 

4.10157 

8.83656 

6.91 
6.92 
6.93 

47.7481 
47.8864 
48.0249 

2.62869 
2.63059 
2.63249 

8.31264 
8.31865 
8.32466 

329.939 
331.374 
332.813 

1.90470 
1.90562 
1.90653 

4.10355 
4.10552 
4.10750 

8.84082 
8.84509 
8.84934 

6.94 
6.95 
6.96 

48.1636 
48.3025 
48.4416 

2.63439 
2.63629 
2.63818 

8.33067 
8.33667 
8.34266 

334.255 
335.702 
337.154 

1.90745 
1.90837 
1.90928 

4.10948 
4.11145 
4.11342 

8.85360 
8.85785 
8.86210 

6.97 
6.98 
6.99 

48.5809 
48.7204 
48.8601 

2.64008 
2.64197 
2.64386 

8.34865 
8.35464 
8.36062 

338.609 
340.068 
341.532 

1.91019 
1.91111 
1.91202 

4.11539 
4.11736 
4.11932 

8.86634 
8.87058 
8.87481 

290 


Powers  and  Boots 


n> 

W2 

V^ 

VWn 

n* 

Vn 

VWn 

^100^ 

7.00 

49.0000 

2.64575 

8.36660 

343.000 

1.91293 

4.12129 

8.87904 

7.01 
7.02 
7.03 

49.1401 
49.2804 
49.4209 

2.64764 
2.64953 
2.65141 

8.37257 
8.37854 
8.38451 

344.472 
345.948 
347.429 

1.91384 
1.91475 
1.91566 

4.12325 
4.12521 
4.12716 

8.88327 
8.88749 
8.89171 

7.04 
7.05 
7.06 

49.5616 
49.7025 
49.8436 

2.65330 
2.65518 
2.65707 

8.39047 
8.39643 
8.40238 

348.914 
350.403 
351.896 

1.91657 
1.91747 
1.91838 

4.12912 
4.13107 
4.13303 

8.89592 
8.90013 
8.90434 

7.07 
7.08 
7.09 

49.9849 
50.1264 
50.2681 

2.65895 
2.66083 
2.66271 

8.40833 
8.41427 
8.42021 

353.393 
354.895 
356.401 

1.91929 
1.92019 
1.92109 

4.13498 
4.13693 

4.13887 

8.90854 
8.91274 
8.91693 

7.10 

50.4100 

2.66458 

8.42615 

357.911 

1.92200 

4.14082 

8.92112 

7.11 
7.12 
7.13 

50.5521 
50.6944 
50.8369 

2.66646 
2.66833 
2.67021 

8.43208 
8.43801 
8.44393 

359.425 
360.944 
362.467 

1.92290 
1.92380 
1.92470 

4.14276 
4.14470 
4.14664 

8.92531 
8.92949 
8.93367 

7.14 
7.15 
7.16 

50.9796 
51.1225 
51.2656 

2.67208 
2.67395 
2.67582 

8.44985 
8.45577 
8.46168 

363.994 
365.526 
367.062 

1.92560 
1.92650 
1.92740 

4.14858 
4.15052 
4.15245 

8.93784 
8.94201 
8.94618 

7.17 

7.18 
7.19 

51.4089 
51.5524 
51.6961 

2.67769 
2.67955 
2.68142 

8.46759 
8.47349 
8.47939 

368.602 
370.146 
371.695 

1.92829 
1.92919 
1.93008 

4.15438 
4.15631 
4.15824 

8.95034 
8.95450 
8.95866 

7.20 

51.8400 

2.68328 

8.48528 

373.248 

1.93098 

4.16017 

8.96281 

7.21 

7.22 
7.23 

51.9841 
52.1284 
52.2729 

2.68514 
2.68701 
2.68887 

8.49117 
8.49706 
8.50294 

374.805 
376.367 
377.933 

1.93187 
1.93277 
1.93366 

4.16209 
4.16402 
4.16594 

8.96696 
8.97110 
8.97524 

7.24 
7.25 
7.26 

52.4176 
52.5625 
52.7076 

2.69072 
2.69258 
2.69444 

8.50882 
8.51469 
8.52056 

379.503 

381.078 
382.657 

1.93455 
1.93544 
1.93633 

4.16786 
4.16978 
4.17169 

8.97938 
8.98351 
8.98764 

7.27 
7.28 
7.29 

62.8529 
52.9984 
53.1441 

2.69629 
2.69815 
2.70000 

8.52643 
8.53229 
8.53815 

384.241 
385.828 
387.420 

1.93722 
1.93810 
1.93899 

4.17361 
4.17552 
4.17743 

8.99176 
8.99588 
9.00000 

7.30 

53.2900 

2.70185 

8.54400 

389.017 

1.93988 

4.17934 

9.00411 

7.31 
7.32 
7.33 

53.4361 
53.5824 
53.7289 

2.70370 
2.70555 
2.70740 

8.54985 
8.55570 
8.56154 

390.618 
392.223 
393.833 

1.94076 
1.94165 
1.94253 

4.18125 
4.18315 
4.18506 

9.00822 
9.01233 
9.01643 

7.34 
7.35 
7.36 

53.8756 
54.0225 
54.1696 

2.70924 
2.71109 
2.71293 

8.56738 
8.57321 
8.57904 

395.447 
397.065 
398.688 

1.94341 
1.94430 
1.94518 

4.18696 
4.18886 
4.19076 

9.02053 
9.02462 
9.02871 

7.37 
7.38 
7.39 

54.3169 
54.4644 
54.6121 

2.71477 
2.71662 

2.71846 

8.58487 
8.59069 
8.59651 

400.316 
401.947 
403.583 

1.94606 
1.94694 
1.94782 

4.19266 
4.19455 
4.19644 

9.03280 
9.03689 
9.04097 

7.40 

54.7600 

2.72029 

8.60233 

405.224 

1.94870 

4.19834 

9.04504 

7.41 
7.42 
7.43 

54.9081 
55.0564 
55.2049 

2.72213 

2.72397 
2.72580 

8.60814 
8.61394 
8.61974 

406.869 
408.518 
410.172 

1.94957 
1.95045 
1.95132 

4.20023 
4.20212 
4.20400 

9.04911 
9.05318 
9.05725 

7.44 
7.45 
7.46 

55.3536 
55.5025 
55.6516 

2.72764 
2.72947 
2.73130 

8.62554 
8.63134 
8.63713 

411.831 
413.494 
415.161 

1.95220 
1.95307 
1.95395 

4.20589 
4.20777 
4.20965 

9.06131 
9.06537 
9.06942 

7.47 
7.48 
7.49 

55.8009 
55.9504 
56.1001 

2.73313 
2.73496 
2.73679 

8.64292 
8.64870 

8.65448 

416.833 

418.509 
420.190 

1.95482 
1.95569 
1.  95656 

4.21153 
4.21341 
4.21529 

9.07347 
9.07752 
9.08156 

Powers  and  Boots 


291 


n 

n2 

Vn 

VlOw 

n8 

Vn 

^10  n 

VlWn 

7.50 

56.2500 

2.73861 

8.66025 

421.875 

1.95743 

4.21716 

9.08560 

7.51 
7.52 
7.53 

56.4001 
56.5504 
56.7009 

2.74044 
2.74226 
2.74408 

8.66603 
8.67179 
8.67756 

423.565 
425.259 
426.958 

1.95830 
1.95917 
1.96004 

4.21904 
422091 
4.22278 

9.08964 
9.09367 
9.09770 

7.54 
7.55 
7.56 

56.8516 
57.0025 
57.1536 

2.74591 
2.74773 
2.74955 

8.68332 
8.68907 
8.69483 

428.661 
430.369 
432.081 

1.96091 
1.96177 
1.96264 

4.22465 
4.22651 

4.22838 

9.10173 
9.10575 
9.10977 

7.57 
7.58 
7.59 

57.3049 
57.4564 
57.6081 

2.75136 
2.75318 
2.75500 

.  8.70057 
8.70632 
8.71206 

433.798 
435.520 
437.245 

1.96350 
1.96437 
1.96523 

4.23024 
4.23210 
4.23396 

9.11378 
9.11779 
9.12180 

7.60 

57.7600 

2.75681 

8.71780 

438.976 

1.96610 

4.23582 

9.12581 

7.61 
7.62 
7.63 

57.9121 
58.0644 
58.2169 

2.75862 
2.76043 
2.76225 

8.72353 
8.72926 
8.73499 

440.711 
442.451 
444.195 

1.96696 
1.96782 
1.96868 

4.23768 
4.23954 
4.24139 

9.12981 
9.13380 
9.13780 

7.64 
7.65 
7.66 

58.3696 
58.5225 
58.6756 

2.76405 
2.76586 
2.76767 

8.74071 
8.74643 
8.75214 

445.944 
447.697 
449.455 

1.96954 
1.97040 
1.97126 

4.24324 
4.24509 
4.24694 

9.14179 

9.14577 
9.14976 

7.67 
7.68 
7.69 

58.8289 
58.9824 
59.1361 

2.76948 
2.77128 
2.77308 

8.75785 
8.76356 
8.76926 

451.218 
452.985 
454.757 

1.97211 
1.97297 
1.97383 

4.24879 
4.25063 
4.25248 

9.15374 
9.15771 
9.16169 

7.70 

59.2900 

2.77489 

8.77496 

456.533 

1.97468 

4.25432 

9.16566 

7.71 
7.72 
7.73 

59.4441 
59.5984 
59.7529 

2.77669 
2.77849 
2.78029 

8.78066 
8.78635 
8.79204 

458.314 
460.100 
461.890 

1.97554 
1.97639 
1.97724 

4.25616 
4.25800 
4.25984 

9.16962 
9.17359 
9.17754 

7.74 
7.75 
7.76 

59.9076 
60.0625 
60.2176 

2.78209 
2.78388 
2.78568 

8.79773 
8.80341 
8.80909 

463.685 
465.484 
467.289 

1.97809 
1.97895 
1.97980 

4.26167 
4.26351 
4.26534 

9.18150 
9.18545 
9.18940 

7.77 
7.78 
7.79 

60.3729 
60.5284 
60.6841 

2.78747 
2.78927 
2.79106 

8.81476 
8.82043 
8.82610 

469.097 
470.911 
472.729 

1.98065 
1.98150 
1.98234 

4.26717 
4.26900 

4.27083 

9.19335 
9.19729 
9.20123 

7.80 

60.8400 

2.79285 

8.83176 

474.552 

1.98319 

4.27266 

9.20516 

7.81 

7.82 
7.83 

60.9961 
61.1524 
61.3089 

2.79464 
2.79643 
2.79821 

8.83742 
8.84308 
8.84873 

476.380 
478.212 
480.049 

1,98404 
1.98489 
1.98573 

4.27448 
4.27631 
4.27813 

9.20910 
9.21302 
9.21695 

7.84 
7.85 
7.86 

61.4656 
61.6225 
61.7796 

2.80000 
2.80179 
2.80357 

8.85438 
8.86002 
8.86566 

481.890 
483.737 
485.588 

1.98658 
1.98742 
1.98826 

4.27995 
4.28177 
4.28359 

9.22087 
9.22479 
9.22871 

7.87 
7.88 
7.89 

61.9369 
62.0944 
62.2521 

2.80535 
2.80713 
2.80891 

8.87130 
8.87694 

8.88257 

487.443 
489.304 
491.169 

1.98911 
1.98995 
1.99079 

4.28540 
4.28722 
4.28903 

9.23262 
9.23653 
9.24043 

7.90 

62.4100 

2.81069 

8.88819 

493.039 

1.99163 

4.29084 

9.24434 

7.91 
7.92 
7.93 

62.5681 
62.7264 
62.8849 

2.81247 
2.81425 
2.81603 

8.89382 
8.89944 
8.90505 

494.914 
496.793 
498.677 

1.99247 
1.99331 
1.99415 

4.29265 
4.29446 
4.29627 

9.24823 
9.25213 
9.25602 

7.94 
7.95 
7.96 

63.0436 
63.2025 
63.3616 

2.81780 
2.81957 
2.82135 

8.91067 
8.91628 
8.92188 

500.566 
502.460 
504.358 

1.99499 
1.99582 
1.99666 

4.29807 
4.29987 
4.30168 

9.25991 
9.26380 
9.26768 

7.97 
7.98 
7.99 

63.5209 
63.6804 
63.8401 

2.82312 
2.82489 
2.82666 

8.92749 
8.93308 
8.93868 

506.262 
508.170 
510.082 

1.99750 
1.99833 
1.99917 

4.30348 
4.30528 
4.30707 

9.27156 
9.27544 
9.27931 

292 


Powers  and  Roots 


n 

W2 

Vii 

VlOn 

W3 

Vn 

VWn 

^lOOn 

8.00 

64.0000 

2.82843 

8.94427 

512.000 

2.00000 

4.30887 

9.28318 

8.01 
8.02 
8.03 

64.1601 
64.3204 
64.4809 

2.83019 
2.83196 
2.83373 

8.94986 
8.95545 
8.96103 

513.922 
515.850 
517.782 

2.00083 
2.00167 
2.00250 

4.31066 
4.31246 
4.31425 

9.28704 
9.29091 
9.29477 

8.04 
8.05 
8.06 

64.6416 
64.8025 
64.9636 

2.83549 
2.83725 
2.83901 

8.96660 
8.97218 
8.97775 

519.718 
521.660 
523.607 

2.00333 
2.00416 
2.00499 

4.31604 
4.31783 
4.31961 

9.29862 
9.30248 
9.30633 

8.07 
8.08 
8.09 

65.1249 
65.2864 
65.4481 

2.84077 
2.84253 
2.84429 

8.98332 
8.98888 
8.99444 

525.558 
527.514 
529.475 

2.00582 
2.00664 
2.00747 

4.32140 
4.32318 
4.32497 

9.31018 
9.31402 
9.31786 

8.10 

65.6100 

2.84605 

9.00000 

531.441 

2.00830 

4.32675 

9.32170 

8.11 
8.12 
8.13 

65.7721 
65.9344 
66.0969 

2-84781 
2.84956 
2.85132 

9.00555 
9.01110 
9.01665 

533.412 
535.387 
537.368 

2.00912 
2.00995 
2.01078 

4.32853 
4.33031 
4.33208 

9.32553 
9.32936 
9.33319 

8.14 
8.15 
8.16 

66.2596 
66.4225 
66.5856 

2.85307 
2.85482 
2.85657 

9.02219 
9.02774 
9.03327 

539.353 
541.343 
543.338 

2.01160 
2.01242 
2.01325 

4.33386 
4.33563 
4.33741 

9.33702 
9.34084 
9.34466 

8.17 
8.18 
8.19 

66.7489 
66.9124 
67.0761 

2.85832 
2.86007 
2.86182 

9.03881 
9.04434 
9.04986 

545.339 
547.343 
549.353 

2.01407 
2.01489 
2.01571 

4.33918 
4.34095 
4.34271 

9.34847 
9.35229 
9.35610 

8.20 

67.2400 

2.86356 

9.05539 

551.368 

2.01653 

4.34448 

9.35990 

8.21 
8.22 
8.23 

67.4041 
67.5684 
67.7329 

2.86531 
2.86705 
2.86880 

9.06091 
9.06642 
9.07193 

553.388 
555.412 
557.442 

2.01735 
2.01817 
2.01899 

4.34625 
4.34801 
4.34977 

9.36370 
9.36751 
9.37130 

8.24 
8.25 
8.26 

67.8976 
68.0625 
68.2276 

2.87054 
2.87228 
2.87402 

9.07744 
9.08295 
9.08845 

559.476 
561.516 
563.560 

2.01980 
2.02062 
2.02144 

4.35153 
4.35329 
4.35505 

9.37510 
9.37889 
9.38268 

8.27 
8.28 
8.29 

68.3929 
68.5584 
68.7241 

2.87576 
2.87750 
2.87924 

9.09395 
9.09945 
9.10494 

565.609 
567.664 
569.723 

2.02225 

2.02307 
2.02388 

4.35681 
4.35856 
4.36032 

9.38646 
9.39024 
9.39402 

8.30 

68.8900 

2.88097 

9.11043 

571.787 

2.02469 

4.36207 

9.39780 

8.31 
8.32 
8.33 

69.0561 
69.2224 
69.3889 

2.88271 
2.88444 
2.88617 

9.11592 
9.12140 
9.12688 

573.856 
575.930 
578.010 

2.02551 
2.02632 
2.02713 

4.36382 
4.36557 
4.36732 

9.40157 
9.40534 
9.40911 

8.34 
8.35 
8.36 

69.5556 
69.7225 
69.8896 

2.88791 
2.88964 
2.89137 

9.13236 
9.13783 
9.14330 

580.094 
582.183 
584.277 

2.02794 
2.02875 
2.02956 

4.36907 
4.37081 
4.37256 

9.41287 
9.41663 
9.42039 

8.37 
8.38 
8.39 

70.0569 
70.2244 
70.3921 

2.89310 
2.89482 
2.89655 

9.14877 
9.15423 
9.15969 

586.376 
588.480 
590.590 

2.03037 
2.03118 
2.03199 

4.37430 
4.37604 
4.37778 

9.42414 

0.42789 
9.43164 

8.40 

70.5600 

2.89828 

9.16515 

592.704 

2.03279 

4.37952 

9.43539 

8.41 
8.42 
8.43 

70.7281 
70.8964 
71.0649 

2.90000 
2.90172 
2.90345 

9.17061 
9.17606 
9.18150 

594.823 
596.948 
599.077 

2.03360 
2.03440 
2.03521 

4.38126 
4.38299 
4.38473 

9.43913 
9.44287 
9.44661 

8.44 
8.45 
8.46 

71.2336 
71.4025 
71.5716 

2.90517 
2.90689 
2.90861 

9.18695 
9.19239 
9.19783 

601.212 
603.351 

605.496 

2.03601 
2.03(582 
2.03762 

4.38646 
4.38819 
4.38992 

9.45034 
9.45407 
9.45780 

8.47 
8.48 
8.49 

71.7409 
71.9104 
72.0801 

2.91033 
2.91204 
2.91376 

9.20326 
9.20869 
9.21412 

607.645 
609.800 
611.960 

2.03842 
2.039L':i 
2.04003 

4.30166 

4.39338 

4.:  s'.>r>  10 

9.46152 
9.46525 
9.46897 

Powers  and  Boots 


293 


n 

w2 

Vti- 

VWn 

n8 

3/n 

VWn 

^100  n 

8.50 

72.2500 

2.91548 

9.21954 

614.125 

2.04083 

4.39683 

9.47268 

8.51 
8.52 
8.53 

72.4201 
72.5904 
72.7609 

2.91719 
2.91890 
2.92062 

9.22497 
9.23038 
9.23580 

616.295 
618.470 
620.650 

2.04163 
2.04243 
2.04323 

4.39855 
4.40028 
4.40200 

9.47640 
9.48011 
9.48381 

8.54 
8.55 
8.56 

72.9316 
73.1025 
73.2736 

2.92233 
2.92404 
2.92575 

9.24121 

9.24662 
9.25203 

622.836 
625.026 
627.222 

2.04402 
2.04482 
2.04562 

4.40372 
4.40543 
4.40715 

9.48752 
9.49122 
9.49492 

.8.57 
8.58 
8.59 

73.4449 
73.6164 

73.7881 

2.92746 
2.92916 
2.93087 

9.25743 
9.26283 
9.26823 

629.423 
631.629 
633.840 

2.04641 
2.04721 
2.04801 

4.40887 
4.41058 
4.41229 

9.498(51 
9.50231 
9.50600 

860 

73.9600 

2.93258 

9.27362 

636.056 

2.04880 

4.41400 

9.50969 

8.61 
8.62 
8.63 

74.1321 
74.3044 
74.4769 

2.93428 
2.93598 
2.93769 

9.27901 
9.28440 
9.28978 

638.277 
640.504 
642.736 

2.04959 
2.03039 
2.05118 

4.41571 
4.41742 
4.41913 

9.51337 
9.51705 
9.52073 

8.G4 
8.65 
8.66 

74.6496 
74.8225 
74.9956 

2.93939 
2.94109 
2.94279 

9.29516 
9.30054 
9.30591 

644.973 
647.215 
649.462 

2.05197 
2.05276 
2.05355 

4.42084 
4.42254 
4.42425 

9.52441 
9.52808 
9.53175 

8.67 
8.68 
8.69 

75.1689 
75.3424 
75.5161 

2.94449 
2.94618 
2.94788 

9.31128 
9.31665 
9.32202 

651.714 
653.972 
656.235 

2.05434 
2.05513 
2.05592 

4.42595 
4.42765 
4.42935 

9.53542 
9.53908 
9.54274 

8.70 

75.0900 

2.<)4958 

9.32738 

658.503 

2.05671 

4.43105 

9.54640 

8.71 
8.72 
8.73 

75.8641 
76.0384 
76.2129 

2.95127 
2.95296 
2.9546(5 

9.33274 
9.33H09 
9.34345 

660.776 
663.055 
665.339 

2.05750 
2.05828 
2.05907 

4.43274 
4.43444 
4.43613 

9.55006 
9.55371 
9.55736 

8.74 
8.75 
8.76 

76.3876 
76.5625 
76.7376 

2.95635 
2.95804 
2.95973 

9.34880 
9.35414 
9.35949 

667.628 
669.922 
672.221 

2.05986 
2.06064 
2.06143 

4.43783 
4.43952 
4.44121 

9.56101 
9.56466 
9.56830 

8.77 
8.78 
8.79 

76.9129 
77.0884 
77.2641 

2.96142 
2.96311 
2.96479 

9.36483 
9.37017 
9.37550 

674.526 
676.836 
679.151 

2.06221 

2.06299 
2.06378 

4.44290 
4.44459 
4.44627 

9.57194 
9.57557 
9.57921 

8  80 

77.4400 

2.96648 

9.38083 

681.472 

2.06456 

4.44796 

9.58284 

8.81 
8.82 
8.83 

77.6161 
77.7924 
77.9689 

2.96816 
2.96985 
2.97153 

9.38616 
9.39149 
9.39681 

683.798 
686.129 
688.465 

2.06534 
2.06612 
2.06690 

4.44964 
4.45133 
4.45301 

9.58647 
9.59009 
9.59372 

8.84 
8.85 
8.86 

78.1456 
78.3225 
78.4996 

2.97321 
2.97489 
2.97658 

9.40213 
9.40744 
9.41276 

690.807 
693.154 
695.506 

2.06768 
2.06846 
2.06924 

4.45469 
4.45637 
4.45805 

9.59734 
9.60095 
9.60457 

8.87 
8.88 
8.89 

78.6769 
78.8544 
79.0321 

2.97825 
2.97993 
2.98161 

9.41807 
9.42338 
9.42868 

697.864 
700.227 
702.595 

2.07002 
2.07080 
2.07157 

4.45972 
4.46140 
4.46307 

9.60818 
9.61179 
9.6154C 

890 

79.2100 

2.98329 

9.43398 

704.969 

2.07235 

4.46475 

9.61900 

8.91 
8.92 
8.93 

79.3881 
79.5664 
79.7449 

2.98496 
2.98664 
2.98831 

9.43928 
9.44458 
9.44987 

707.348 
709.732 
712.122 

2.07313 
2.07390 
2.07468 

4.46642 
4.46809 
4.46976 

9.62260 
9.62620 
9.62980 

8.94 
8.95 
8.96 

79.9236 
80.1025 
80.2816 

2.98998 
2.99166 
2.99333 

9.45516 
9.46044 
9.46573 

714.517 
716.917 
719.323 

2.07545 
2.07622 
2.07700 

4.47142 
4.47309 
4.47476 

9.63339 
9.63698 
9.64057 

8.97 
8.98 
8.99 

80.4609 
80.6404 
80.8201 

2.99500 
2.99666 
2.99833 

9.47101 
9.47629 
9.48156 

721.734 
724.151 
726.573 

2.07777 
2.07854 
2.07931 

4.47642 
4.47808 
4.47974 

9.64415 
9.64774 
9.65132 

294 


Powers  and  Boots 


n 

n* 

Vti 

VWn 

n* 

V* 

VWn 

VJMn 

9.00 

81.0000 

3.00000 

9.48683 

729.000 

2.08008 

4.48140 

9.65489 

9.01 
9.02 
9.03 

81.1801 
81.3604 
81.5409 

3.00167 
3.00333 
3.00500 

9.49210 
9.49737 
9.50263 

731.433 
733.871 
736.314 

2.08085 
2.08162 
2.08239 

4.48306 
4.48472 
4.48638 

9.65847 
9.66204 
9.66561 

9.04 
9.05 
9.06 

81.7216 
81.9025 
82.0836 

3.00666 
3.00832 
3.00998 

9.50789 
9.51315 
9.51840 

738.763 
741.218 
743.677 

2.08316 
2.08393 
2.08470 

4.48803 
4.48969 
4.49134 

9.66918 
9.67274 
9.67630 

9.07 
9.08 
9.09 

82.2649 
82.4464 
82.6281 

3.01164 
3.01330 
3.01496 

9.523(35 
9.52890 
9.53415 

746.143 

748.613 
751.089 

2.08546 
2.08623 
2.08699 

4.49299 
4.49464 
4.49629 

9.67986 
9.68342 
9.68697 

9.10 

82.8100 

3.01662 

9.53939 

753.571 

2.08776 

4.49794 

9.69052 

9.11 
9.12 
9.13 

82.9921 
83.1744 
83.3569 

3.01828 
3.01993 
3.02159 

9.54463 
9.54987 
9.55510 

756.058 
758.551 
761.048 

2.08852 
2.08929 
2.09005 

4.49959 
4.50123 
4.50288 

9.69407 
9.69762 
9.70116 

9.14 
9.15 
9.16 

83.5396 
83.7225 
83.9056 

3.02324 
3.02490 
3.02655 

9.56033 
9.56556 
9.57079 

763.552 
766.061 
768.575 

2.09081 
2.09158 
2.09234 

4.50452 
4.50616 
4.50781 

9.70470 
9.70824 
9.71177 

9.17 
8.18 
9.19 

84.0889 
84.2724 
84.4561 

3.02820 
3.02985 
3.03150 

9.57601 
9.58123 

9.58645 

771.095 
773.621 
776.152 

2.09310 

2.09386 
2.09462 

4.50945 
4.51108 
4.51272 

9.71531 
9.71884 
9.72236 

9.20 

84.6400 

3.03315 

9.59166 

778.688 

2.09538 

4.51436 

9.72589 

9.21 
9.22 
9.23 

84.8241 
85.0084 
85.1929 

3.03480 
3.03645 
3.03809 

9.59687 
9.60208 
9.60729 

781.230 
783.777 
786.330 

2.09614 
2.09690 
2.09765 

4.51599 
4.51763 
4.51926 

9.72941 
9.73293 
9.73645 

9.24 
9.25 
9.26 

85.3776 
85.5625 
85.7476 

3.03974 
3.04138 
3.04302 

9.61249 
9.61769 
9.62289 

788.889 
791.453 
794.023 

2.09841 
2.09917 
2.09992 

4.52089 
4.52252 
4.52415 

9.73996 
9.74348 
9.74699 

9.27 

9.28 
9.29 

85.9329 
86.1184 
86.3041 

3.04467 
3.04631 
3.04795 

9.62808 
9.63328 
9.63846 

796.598 
799.179 
801.765 

2.10068 
2.10144 
2.10219 

4.52578 
4.52740 
4.52903 

9.75049 
9.75400 
9.75750 

9.30 

86.4900 

3.04959 

9.64365 

804.357 

2.10294 

4.53065 

9.76100 

9.31 
9.32 
9.33 

86.6761 
86.8624 
87.0489 

3.05123 
3.05287 
3.05450 

9.64883 
9.65401 
9.65919 

806.954 
809.558 
812.166 

2.10370 
2.10445 
2.10520 

4.53228 
4.53390 
4.53552 

9.76450 
9.76799 
9.77148 

9.34 
9.35 
9.36 

87.2356 
87.4225 
87.6096 

3.05614 
3.05778 
3.05941 

9.66437 
9.66954 
9.67471 

814.781 
817.400 
820.026 

2.10595 
2.10671 
2.10746 

4.53714 
4.53876 
4.54038 

9.77497 
9.77846 
9.78195 

9.37 
9.38 
9.39 

87.7969 
87.9844 
88.1721 

3.06105 
3.06268 
3.06431 

9.67988 
9.68504 
9.69020 

822.657 
825.294 
827.936 

2.10821 
2.10896 
2.10971 

4.54199 
4.54361 
4.54522 

9.78543 
9.78891 
9.79239 

9.40 

88.3600 

3.06594 

9.69536 

830.584 

2.11045 

4.54684 

9.79586 

9.41 
9.42 
9.43 

88.5481 
88.7364 
88.9249 

3.06757 
3.06920 
3.07083 

9.70052 
9.70567 
9.71082 

833.238 
835.897 
838.562 

2.11120 
2.11195 
2.11270 

4.54845 
4.55006 
4.55167 

9.79933 
9.80280 
9.80627 

9.44 
9.45 
9.46 

89.1136 
89.3025 
89.4916 

3.07246 
3.07409 
3.07571 

9.71597 
9.72111 
9.72625 

841.232 
843.909 
846.591 

2.11344 
2.11419 
2.11494 

4.55328 
4.55488 
4.55649 

9.80974 
9.81320 
9.81666 

9.47 
9.48 
9.49 

89.6809 
89.8704 
90.0601 

3.07734 
3.07896 
3.08058 

9.73139 
9.73653 
9.74166 

849.278 
851.971 

854.670 

2.11568 
2.11642 
2.11717 

4.55809 
4.55970 
4.56130 

9.82012 
9.82357 
9.82703 

Powers  and  Roots 


295 


n 

n2 

Vn 

VWn 

n8 

^n 

^10  n 

^100  n 

9.50 

90.2500 

3.08221 

9.74679 

857.375 

2.11791 

4.56290 

9.83048 

9.51 
9.52 
9.53 

90.4401 
90.6304 
90.8209 

3.08383 
3.08545 
3.08707 

9.75192 
9.75705 
9.76217 

860.085 
862.801 
865.523 

2.11865 
2.11940 
2.12014 

4.56450 
4.56610 
4.56770 

9.83392 
9.83737 
9.84081 

9.54 
9.55 
9.56 

91.0116 
91.2025 
91.3936 

3.08869 
3.09031 
3.09192 

9.76729 
9.77241 
9.77753 

868.251 
870.984 
873.723 

2.12088 
2.12162 
2.12236 

4.56930 
4.57089 
4.57249 

9.84425 
9.84769 
9.85113 

9.57 
9.58 
9.59 

91.5849 
91.7764 
91.9681 

3.09354 
3.09516 
3.09677 

9.78264 
9.78775 
9.79285 

876.467 
879.218 
881.974 

2.12310 
2.12384 
2.12458 

4.57408 
4.57567 

4.57727 

9.85456 
9.85799 
9.86142 

9.60 

92.1600 

3.09839 

9.79796 

884.736 

2.12532 

4.57886 

9.86485 

9.61 
9.62 
9.63 

92.3521 
92.5444 
92.7369 

3.10000 
3.10161 
3.10322 

9.80306 
9.80816 
9.81326 

887.504 
890.277 
893.056 

2.12605 
2.12679 
2.12753 

4.58045 
4.58204 
4.58362 

9.86827 
9.87169 
9.87511 

9.64 
9.65 
9.66 

92.9296 
93.1225 
93.3156 

3.10483 
3.10644 
3.10805 

9.81835 
9.82344 
9.82853 

895.841 
898.632 
901.429 

2.12826 
2.12900 
2.12974 

4.58521 
4.58679 
4.58838 

9.87853 
9.88195 
9:88536 

9.67 
9.68 
9.69 

93.5089 
93.7024 
93.8961 

3.10966 
3.11127 
3.11288 

9.83362 
9.83870 
9.84378 

904.231 
907.039 
909.853 

2.13047 
2.13120 
2.13194 

4.58996 
4.59154 
4.59312 

9.88877 
9.89217 
9.89558 

9.70 

94.0900 

3.11448 

9.84886 

912.673 

2.13267 

4.59470 

9.89898 

9.71 
9.72 
9.73 

94.2841 
94.4784 
94.6729 

3.11609 
3.11769 
3.11929 

9.85393 
9.85901 
9.86408 

915.499 
918.330 
921.167 

2.13340 
2.13414 
2.13487 

4.59628 
4.59786 
4.59943 

9.90238 
9.90578 
9.90918 

9.74 
9.75 
9.76 

94.8676 
95.0625 
95.2576 

3.12090 
3.12250 
3.12410 

9.86914 
9.87421 
9.87927 

924.010 

926.859 
929.714 

2.13560 
2.13633 
2.13706 

4.60101 
4.60258 
4.60416 

9.91257 
9.91596 
9.91935 

9.77 
9.78 
9.79 

95.4529 
95.6484 
95.8441 

3.12570 
3.12730 
3.12890 

9.88433 
9.88939 
9.89444 

932.575 
935.441 
938.314 

2.13779 
2.13852 
2.13925 

4.60573 
4.60730 

4.60887 

9.92274 
9.92612 
9.92950 

9.80 

96.0400 

3.13050 

9.89949 

941.192 

2.13997 

4.61044 

9.93288 

9.81 
9.82 
9.83 

96.2361 
96.4324 
96.6289 

3.13209 
3.13369 
3.13528 

9.90454 
9.90959 
9.91464 

944.076 
946.966 
949.862 

2.14070 
2.14143 
2.14216 

4.61200 
4.61357 
4.61514 

9.93626 
9.93964 
9.94301 

9.84 
9.85 
9.86 

96.8256 
97.0225 
97.2196 

3.13688 
3.13847 
3.14006 

9.91968 
9.92472 
9.92975 

952.764 
955.672 
958.585 

2.14288 
2.14361 
2.14433 

4.61670 
4.61826 
4.61983 

9.94638 
9.94975 
9.95311 

9.87 
9.88 
9.89 

97.4169 

97.  (5144 
97.8121 

3.14166 
3.14325 
3.14484 

9.93479 
9.93982 
9.94485 

961.505 
964.430 
967.362 

2.14506 
2.14578 
2.14651 

4.62139 
4.62295 
4.62451 

9.95648 
9.95984 
9.96320 

9.90 

98.0100 

3.14643 

9.94987 

970.299 

2.14723 

4.62607 

9.96655 

9.91 
9.92 
9.93 

98.2081 
98.4064 
98.6049 

3.14802 
3.14960 
3.15119 

9.95490 
9.95992 
9.96494 

973.242 
976.191 
979.147 

2.14795 
2.14867 
2.14940 

4.62762 
4.62918 
4.63073 

9.96991 
9.97326 
9.97661 

9.94 
9.95 
9.96 

98.8036 
99.0025 
99.2016 

3.15278 
3.15436 
3.15595 

9.96995 
9.97497 
9.97998 

982.108 
985.075 
988.048 

2.15012 
2.15084 
2.15156 

4.63229 
4.63384 
4.63539 

9.97996 
9.98331 
9.98665 

9.97 
9.98 
9.99 

99.4009 
99.6004 
99.8001 

3.15753 
3.15911 
3.16070 

9.98499 
9.98999 
9.99500 

991.027 
994.012 
997.003 

2.15228 
2.15300 
2.15372 

4.63694 
4.63849 
4.64004 

9.98999 
9.99333 

9.99667 

TABLE   II  — IMPORTANT  NUMBERS 

A.    Units  of  Length 
ENGLISH  UNITS  METRIC  UNITS 

12  inches  (in.)  =  1  foot  (ft.)          10  millimeters  =  1  centimeter  (cm.) 
3  feet  =  1  yard  (yd.)  (mm.) 

6£  yards          =  1  rod  (rd.)  10  centimeters  =  1  decimeter  (dm.) 

320 rods  =  1  mile  (mi.)        10  decimeters   =  1  meter  (m.) 

10  meters          =  1  dekameter  (Dm.) 
1000  meters      =  1  kilometer  (Km.) 

ENGLISH  TO  METRIC  METRIC  TO  ENGLISH 

1  in.  =  2.5400  cm.  1  cm.   =  0.3937  in. 

1  ft.    =  30.480  cm.  1m.     =  39.37  in.  =  3.2808  ft 

1  mi.  =  1.6093  Km.  1  Km.  =  0.6214  mi. 

B.    Units  of  Area  or  Surface 

1  square  yard  =  9  square  feet  =  1296  square  inches 
1  acre  (A.)  =  160  square  rods  =  4840  square  yards 
1  square  mile  =  640  acres  =  102400  square  rods 

C.    Units  of  Measurement  of  Capacity 

DRY  MEASURE  LIQUID  MEASURE 

2  pints  (pt.)  =  1  quart  (qt.)  4  gills  (gi.)  =  1  pint  (pt.) 

8  quarts         =  1  peck  (pk.)  2  pints         =  1  quart  (qt.) 

4  pecks          =  1  bushel  (bu.)  4  quarts       =  1  gallon  (gal.) 

1  gallon       =  231  cu.  in. 

D.    Metric  Units  to  English  Units 

1  liter  =  1000  cu.  cm.  =  61.02  cu.  in.  =  1.0567  liquid  quarts 
1  quart  =  .94636  liter  =  946.36  cu.  cm. 
1000  grams  =  1  kilogram  (Kg.)  =  2.2046  pounds  (Ib.) 
1  pound  =  .453593  kilogram  =  453.59  grams 

E.    Other  Numbers 

T  =  ratio  of  circumference  to  diameter  of  a  circle 

=  3.14159265 
1  radian  =  angle  subtended  by  an  arc  equal  to  the  radius 

=  57°  17'  44".8  =  67°.2957795  =  180°/ir 
1  degree  =  0.01745329  radian,  or  w/lSQ  radians 
Weight  of  1  cu.  ft.  of  water  =  62.425  Jb. 
296 


INDEX 


Abscissa,  41. 

Absolute  value,  2. 

Addition,  of  expressions,  9;  of  frac- 
tions, 32 ;  of  radicals,  74. 

Antecedent,  166. 

Arithmetic  progression,  141 ;  means, 
146. 

Ascending  powers,  8. 

Axes,  coordinate,  41. 

Base,  logarithm  to  any,  242. 
Brace,   9. 
Bracket,  9. 

Binomial,  8;  theorem,  258;  proof 
of  theorem,  261. 

Calculating  machines,  244. 
Change  of  signs  in  fractions,  29. 
Characteristic,  219,  221,  223. 
Coefficient,  8. 
Common  difference,   141. 
Common  logarithm,   242. 
Complex  numbers,  93. 
Consequent,   166. 
Constants,  181. 
Coordinates  of  a  point,  41. 
Cube  root,  5. 

Decimals,  repeating,  161. 
Denominator,   29. 
Descending  powers,  8. 
Determinant,    of   the   second    order, 

264;   of  the  third   order,  269;    of 

higher  order,  273. 
Difference,   tabular,  230. 
Discriminant,    110. 
Division,  formulas  and  rules,  14. 

Elements  of  a  determinant,  265. 
Elimination     by     substitution,     50; 

by  addition  or  subtraction,  51. 
Ellipse,   125. 


Equation,  simple,  36 ;  linear,  36 ; 
of  the  first  degree,  36;  solution 
of,  36;  root  of,  36;  principles 
useful  in  the  solution  of,  37,  38; 
containing  radicals,  65;  literal, 
95;  quadratic,  see  Quadratic 
equation. 

Equations,  simultaneous,  47;  in- 
consistent, 48;  simultaneous  in 
three  unknowns,  54. 

Evolution,    196. 

Exponent,  4;  fractional,  198;  zero, 
199 ;  fundamental  laws  for  any 
rational  exponent,  201. 

Exponents,  laws  of,  193 ;  introduc- 
tion of  general,  198;  negative  in 
fractions,  200. 

Extremes  of  a  proportion,    167. 

Factor,  prime,  26. 

Factoring,  type  forms  of,  19,  23. 

Factors,  common,  26 ;  highest  com- 
mon, 26. 

Formulas,  97,   102,  205. 

Fractions,  definition,  28 ;  equiva- 
lent, 29 ;  change  of  signs  in,  29 ; 
reduction  to  lowest  terms,  30; 
reduction  to  lowest  common  de- 
nominator, 31 ;  addition  and  sub- 
traction of,  32 ;  multiplication 
and  division  of,  36. 

Functions,  idea  of,  245 ;  types  of  alge- 
braic functions,  246;  considered 
graphically,  250. 

Gear  wheel  law,  101. 

Geometric  progression,  150;  means, 
154;  infinite,  156. 

Graph,  of  an  equation,  43 ;  deter- 
mined from  two  points,  44 ;  of  a 
quadratic,  90 ;  of  a  function,  250. 


297 


298 


INDEX 


Hyperbola,  126. 

Imaginary  numbers,  116;  pure, 
93;  unit,  116;  addition  and  sub- 
traction of,  118;  multiplication 
of,  119;  division  of,  120;  geo- 
metric representation  of,  121. 

Inconsistent  equations,  47. 

Index,  5,  68,  196. 

Involution,  193. 

Irrational  numbers,  68. 

Letters,  use  of  in  algebra,  4. 

Lever,  law  of,   100. 

Limit,  variable,   159. 

Linear  equation,  36 ;  graph  of,  43,  44. 

Logarithm,  217 ;  of  any  number, 
217 ;  number  corresponding  to, 
231;  of  a  power,  235;  general, 
242;  common,  242;  of  a  root, 
244;  tables,  291. 

Mantissa,  219;  determination  of, 
225. 

Mathematical  induction,   255. 

Mean  proportional,  172. 

Means  of  a  proportion,   167. 

Monomial,    8 ;  square  root  of,  62. 

Multiple,  common,  27 ;  lowest  com- 
mon, 28. 

Multiplication,  formulas  and  rules, 
11. 

Negative  exponents,  199. 

Negative    numbers,    1 ;     operations 

with,  2. 
Numbers,  negative,  1 ;    positive,  1 ; 

operations  with,  2;    rational,   68, 

108;   irrational,  68;    complex,  93; 

real,    93,     108;      summary,     109; 

imaginary,  see  Imaginary  number. 
Numerator,  29. 

Order  of  a  determinant,  265,  269. 
Ordinate,  41. 
Origin,   41. 

Parenthesis,   9. 

Perfect  square  trinomial,  20. 


Periods  in  square  root,  60. 

Polynomial,  8;  arranging  a,  8; 
square  root  of,  62. 

Power,  4;    tables,   274. 

Powers,  195. 

Prime    factor,    27. 

Progression,  arithmetic,  141 ;  geo- 
metric, 150. 

Proportion,  167;  terms  of  a,  167; 
extremes  of  a,  167 ;  means  of  a, 
167;  algebraic,  168;  fundamental 
principles  of,  168,  174;  inversion 
in  a,  174 ;  alternation  in  a,  174 ; 
composition  in  a,  174 ;  division 
in  a,  175;  composition  and  divi- 
sion in  a,  175 ;  several  equal  ratios 
in  a,  176. 

Proportional,  mean,  172;  third  and 
fourth,  172. 

Quadratic  equation,  78;  pure,  78; 
affected,  78;  solution  of  pure, 
78 ;  solution  of  affected  by  fac- 
toring, 81 ;  solution  by  complet- 
ing the  square,  83,  85;  solution 
by  the  Hindu  method,  86 ;  solu- 
tion by  formula,  87 ;  graphical 
solution  of,  90,  122 ;  having  imag- 
inary solutions,  92;  character  of 
the  roots  of,  109 ;  character  of  roots 
considered  geometrically,  111;  for- 
mation of  from  given  solutions, 
114;  solution  by  elimination, 
128;  simultaneous,  137. 

Radical,  or  quadratic  radical,  67; 
of  the  nth  order,  68 ;  value  of, 
68. 

Radicals,  simplification!  of,  71,  206; 
similar,  74;  addition  and  sub- 
traction of,  74;  multiplication 
of,  75;  division  of,  77. 

Radicand,  68,  196. 

Ratio,  166;  of  geometric  progres- 
sion, 150. 

Rational  number,  68. 

Rationalizing  the  denominator,  2J.1. 

Root,  square,  5;  nth,  5,  196. 

Root  of  an  equation,  36. 


INDEX 


299 


Roots,    196;    imaginary  of  a  quad- 
ratic, 108;  character  of,  111 

Simultaneous  equations,   47. 
Slide  rule,  244. 
Solution  of  an  equation,  44. 
Special  products,  formulas  of,  17. 
Square   root,  5 ;    of   a   number,  59 ; 

in  arithmetic,  59 ;    in  algebra,  62 ; 

of      trinomials,    62 ;    double    sign 

of,  64 ;    tables,  274. 
Subtraction,    8 ;     of   expressions,    9 ; 

of  fractions,  32;    of  radicals,  74. 
Surd,  68;    binomial,  214. 


Table,  use  of,  68. 
Term,  8. 

Terms,  like,  8 ;   of  a  proportion,  167. 
Theorem,  binomial,  258. 
Trinomial,    8;     perfect    square,    20; 
square  root  of,  62. 

Unit,   imaginary,    116. 

Variables,    181. 

Variation,  direct,  178;    inverse,  179; 

joint,     180;      problems    in,     185; 

geometrically  considered,  190. 
Vinculum,  9. 


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